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AN 



ELEMENTARY TREATISE 



MECHANICS, 



EMBRACING THE 



THEORY OF STATICS AND DYNAMICS, 



AND ITS APPLICATION TO 



SOLIDS AND FLUIDS. 



^repareti for tlje SStrtJer^raTmate Course fn tlje ffiStesIegan ^mbersfts. 



BY AUGUSTUS W.'SMITH, 1LD., 

PROFESSOR OF MATHEMATICS AND ASTRONOMY. 



NEW YORK: 

HARPER & BROTHERS, PUBLISHERS, 

82 CLIFF STREET. 

1849. 



Entered, according to Act of Congress, in the year one thousand 
eight hundred and forty-nine, by 

Harper & Brothers, 

in the Clerk's Office of the District Court of the Southern District 
of New York. 



/ 



PREFACE. 



The preparation of the present treatise was undertaken un- 
der the impression that an elementary work on analytical Me- 
chanics, suited to the purposes and exigencies of the course of 
study in colleges, was needed. This impression is the result 
of long experience in teaching, and a fair trial of all known 
American works and reprints designed for such use. 

It can scarcely be necessary, at this period, to assign at 
length the reasons for adopting the analytical methods of in- 
vestigation. Whether the object be intellectual discipline, or 
a knowledge of facts and principles, or both, the preference 
must be given to the modern analysis. It affords a wider field 
for the exercise of judgment, calls more fully into exercise the 
inventive powers, and taxes the memory less with unimportant 
particulars, thus developing and strengthening more of the 
mental faculties, and more equably by far than the geometric- 
al methods. It is more universal in its application, shorter in 
practice, and far more fruitful in results. It is, indeed, the only 
method by which the student can advance beyond the merest 
rudiments of the science, without an expense of time and en- 
ergy wholly disproportioned to the ends accomplished — the 
only method by which he can acquire a self-sustaining and a 
progressive power. 

As the hope of furnishing to the student some additional 
facilities for a pleasant and profitable prosecution of this 
branch of study was the motive for undertaking the prepara- 
tion of this manual, it will be proper to refer to some of the 

2 



PREFACE. 



specific objects had in view. The most formidable obstacles 
to the acquisition of any branch of science are generally found 
at the very outset. It has therefore been a specific object to 
introduce the subject by giving distinctness to the elementary 
truths, dwelling upon them till they are rendered familiar, 
adopting the simplest mode of investigation and proof consist- 
ent with rigor of demonstration, and avoiding all reference to 
the metaphysics of the science as out of place in a work de- 
signed for beginners. At every successive stage of advance- 
ment the student is required to review the ground passed over 
by the use of the principles learned, in the solution of exam- 
ples which will require their application, and test at once his 
knowledge of them and his ability to apply them. Having 
examined in this way each division of forces as classified, 
more general methods are introduced, and their application 
illustrated by numerous problems. These methods are often 
employed in particular cases when others less general would 
be shorter ; for it is a readiness in their use and a familiarity 
with their application that gives to the student his power over 
more difficult and complicated questions. 

To adapt the work to the exigencies of the recitation-room, 
the whole is divided, at the risk, perhaps, of too much apparent 
formality, into distinct portions or propositions, suitable for an 
individual exercise. In each case the object to be accomplish- 
ed is distinctly proposed, or the truth to be established is 
briefly and clearly enunciated. The student has thus a defin- 
ite object before him when called upon to recite, acquires a 
convenient formulary of words by which to quote and apply 
his arguments, and the more clearly conceives and marks his 
own progress. 

In this institution the mathematical and experimental courses 
are assigned to different departments. It is not, however, for 
this reason alone that I have purposely abstained from swell- 



PREFACE. 1U 

ing the volume with diffuse verbal explanations, and the intro- 
duction of experimental illustrations. The experienced teach- 
er will always have at command an abundance of matter of 
this kind to meet every emergency, and can adapt the kind 
and mode of illustration to the specific difficulty that arises in 
the mind of his pupil. There is an objection to the introduc- 
tion of such matter, growing out of a tendency, in some at 
least, to become the passive recipients of their mental aliment. 
These mental dyspeptics loathe that which costs them labor, 
and, satisfied with the stimulus of inflation, seize upon the 
lighter portions, and neglect that which alone can impart 
vigor to their mental constitution. Whoever caters for this 
class of students must share in the responsibility of raising up 
a race effeminate in mind, if not in body. 

The investigations are limited to forces in the same plane, 
except in the case of parallel forces. This limit, while it is 
sufficiently ample to include the most important topics of ter- 
restrial mechanics, and to embrace an interesting field of celes- 
tial mechanics, is fully sufficient to occupy the time which the 
crowded course of study in our colleges will admit of being 
appropriated to it. It may, without marring the integrity of 
the work, be very much reduced by the omission of all those 
portions in which the integral calculus is employed ; and still 
further, if desired, by omitting all that relates to the principle 
of virtual velocities, and its application to the mechanical 
powers. A copious analysis is given in the contents, conven- 
ient for frequent and rapid reviews, and suggestive of ques- 
tions for examination. 

In this work no claim is advanced to originality. The ma- 
terials have been sought and freely taken from all available 
sources. Nearly all the matter, in some form, is found in al- 
most every author consulted, and credit could not be given in 
every case, if, in an elementary work designed as a text-book, 



IV PREFACE. 

it were desirable to do so. In the portion on Statics, I am 
most indebted to the excellent introductory work of Profess- 
or Potter, of University College, London. The chapter on 
Couples is substantially taken from Poinsot's Elemens de Sta- 
tique. The questions, generally simplified in their character, 
are mostly taken from Walton's Mechanical and Hydrostatical 
Problems, and Wrigley and Johnston's Examples. The works 
more especially consulted are those of Poisson, Francaeur, 
Gregory, Whewell, Walker, Moseley, and Jamieson. Some- 
thing has been taken from each, but modified to suit the spe- 
cific object kept constantly in view — the preparation of a 
manual which should be simple in its character, would most 
naturally, easily, and successfully induct the student into the 
elementary principles of the science, and prepare him, if so 
disposed, to prosecute the study further, without the necessity 
of beginning again and studying entirely new methods. How 
far I have succeeded must be left to the decision of others, 
especially of my co-laborers in this department of instruction. 

Wesleyan University, 
Middletown, Conn., Jan., 1849. 



CONTENTS. 



Art Page 

INTRODUCTION 1 

1. Definition of Mechanics and its Subdivisions. 

2. Definition of Force and its Mechanical Effects. 

3. Definition of its Intensity and Measure, its Direction and Point of Application. 

4. Definition of Analytical Mechanics. 

5. Definition of Concurring and Conspiring Forces. 

6. Definition of Body, Rigid, Flexible, and Elastic. 



STATICS. 

CHAPTER I. 

COMPOSITION AND EQUILIBRIUM OF CONCURRING FORCES IN THE SAME PLANE. 3 

7. Two equal and opposite Forces in Equilibrium. 

8. Two Forces inclined to each other can not Equilibrate. 

9. Definition of Resultant and Components. 

10. Resultant of several Conspiring Forces. 

11. Resultant of two unequal opposite Forces. 
12 Resultant of any number of opposite Forces. 

13. Point of Application at any Point in its Direction. 

14. Direction of the Resultant of several Forces. 

15. Direction of the Resultant of two equal Forces. 

16. Direction of the Resultant of two unequal Forces. 

17. Variation of the Magnitude of the Resultant and its Components. 

18. Equilibrium of three equal Forces. 

19. Resultant of two equal Forces at an Angle of 120°. 

20. Each of three equilibrating Forces equal to the Resultant of the other two in 

Magnitude. 

21. Parallelogram of Forces. 

22. Triangle of Forces. 

23. Resolution of Forces. 

24. Polygon of Forces. 

25. Representation of equilibrated Forces. 

26. Graphical determination of Resultant. 

27. Parallelopiped of Forces. 

28. Ratios of three equilibrated Forces. 

29. Expression for the Resultant of two Forces. 

30. Definition of Moment of a Force — Origin of Moments. 

31. Equality of the Moment of the Resultant with the Sum of the Moments of two 

Components. 

32. Equality of the Moment of the Resultant with the Sum of the Moments of any 

number of Components. 

33. When the Origin of Moments is fixed. 

34. When the Forces are in Equilibrium. 

35. Examples. 



VI CONTENTS. 

Art. Page 

CHAPTER II. 

PARALLEL FORCES 15 

36. Resultant of two Parallel Forces. 

37. Definition of Arms of Forces. 

38. Equilibrium of two Parallel Forces by a third Force. 

39. Point of Application of the Resultant. 

40. When the Forces act in opposite Directions. 

41. When the Forces ai*e Equal and Opposite. 

42. Such Forces constitute a Statical Couple. 

43. Resultant of any Number of Parallel Forces. 

44. Definition of Center of Parallel Forces. 

45. Equality of the Moment of the Resultant with the Sum of the Moments of the 

Components. 

46. Definition of Moment of a Force in reference to a Plane. 

47. Conditions of Equilibrium of any Number of Parallel Forces. 

48. Condition of Rotation. 

49. When in Equilibrium, each Force equal in Magnitude to the Resultant of all the 

others. 

50. Equilibrium independent of their Direction. 

51. Examples. 

CHAPTER III. 

THEORY OF COUPLES 24 

52. Definition of a Statical Couple. 

53. Definition of the Arms of a Couple. 

54. Definition of the Moment of a Couple. 

55. A Couple may be turned round in its own Plane. 

56. A Couple may be removed parallel to itself in its own Plane. 

57. A Couple may be removed to a Parallel Plane. 

58. Couples are equivalent when their Planes are Parallel and Moments are Equal. 

59. Couples may be changed into others having Arms of a given Length. 

60. Definition of the Axis of a Couple. 

61. Properties of an Axis. 

62. Definition of the Resultant of two or more Couples. 

63. Equality of the Moment of the Resultant with the Sam of the Moments of the 

Components. 

64. Equality of the Axis of the Resultant with the Sum of tine Axes of the Com- 

ponents. 

65. The Resultant of two Couples inclined to each other. 

66. Representative of the Axis of the Resultant of two Couples. 

67. Parallelogram of Couples. 

CHAPTER IV. 

ANALYTICAL STATICS IN TWO DIMENSIONS 31 

68. Resultant of any Number of Concurring Forces. 

69. Directions of the Rectangular Components involved in their Trigonometrical Values. 

70. Conditions of Equilibrium of Concurring Forces. 

71. Resultant Force and Resultant Couple when the Forces do not concur. 

72. Construction of the Results. 

73. Equation of the Resultant. 

74. Equilibrium of non-concurring Forces. 

75. Equilibrium when there is a fixed Point in the System. 

76. Equilibrium of a Point on a Plane Curve, 



CONTENTS. Vll 

Art. Page 

77. Conditions of Equilibrium. 

78. Definition of Virtual Velocities. 

79. Principle of Virtual Velocities. 

80. Principle of Virtual Velocities obtains in Concurring' Forces in the same Plane. 

81. Principle of Virtual Velocities obtains in non-concurring Forces in the same Plane. 

82. The Converse. 

CHAPTER V. 

CENTER OF GRAVITY 43 

83. Definition of Gravity. 

84. Laws of Gravity. 

85. Definition of a Heavy Body. 

B6. Definition of the Weight of a Body. 

87. Definition of its Mass. 

88. Expression for Weight. 

89. Definition of Density. 

90. Another Expression for Weight. 

91. Relations of Masses to Volumes of the same Density. 

92. Relations of Densities to Volumes of the same Mass. 

93. Relations of Densities to Masses of the same Volume. 

94. Definition of Center of Gravity. 

95. Connection of the Center of Gravity with the Doctrine of Parallel Forces. 

96. Definition of a Body symmetrical with respect to a Plane. 

97. Position of its Center of Gravity. 

98. Definition of a Body symmetrical with respect to an Axis. 

99. Position of its Center of Gravity. 

100. Center of Gravity of a Body symmetrical with respect to two Axes. 

101. Definition of Center of Figure. 

102. Center of Gravity of any Number of heavy Particles. 

103. Their Center of Gravity when their Positions are given by their Co-ordinates. 

104. Their ^Center of Gravity when they are all in the same Line. 

105. Their Center of Gravity when they are Homogeneous. 

106. The Center of Gravity of the Whole and a Part given to find that of the other Part. 

107. Examples — 1. Of a Straight Line — 2. Triangle— 3. Parallelogram — 4. Polygon — 

5. Triangular Pyramid — 6. Any Pyramid — 7. Frustum of a Cone — 8. Perimeter 
of a Triangle — 9. Of a Triangle in Terms of its Co-ordinates. 

CONDITIONS OF EQUILIBRIUM OF BODIES FROM THE ACTION OF GRAVITY... 54 

108. When the Body has a Fixed Point in it. 

109. Definition of Stable, Unstable, and Neutral Equilibrium. 

110. Position of the Center of Gravity when the Equilibrium is Stable and when Unstable. 

111. Pressure on the Fixed Point. 

112. Position and Pressure when there are two Fixed Points. 

113. Position and Pressure when there are three Fixed Points. 

114. Position and Pressure when a Body touches a Hoi-izontal Plane in one Point. 

115. Position and Pressure when a Body touches a Horizontal Plane in two Points, i 

116. Position and Pressure when a Body touches a Horizontal Plane in three Points. 

117. Position and Pressure when a Body touches a Horizontal Plane in any Number of 

Points. 

118. Measure of the Stability on a Horizontal. Plane. 

119. Case of a Body on an Inclined Plane. 

120. Examples. 

APPLICATION OF THE INTEGRAL CALCULUS TO THE DETERMINATION OF THE CEN- 
TER OF GRAVITY .59 



Vlll CONTENTS. 

Art. Page 

121. General differential Expressions for the Co-ordinates of the Center of Gravity. 

122. General differential Expx-essions for the Center of Gravity of a Plane Curve. 

123. General differential Expressions for the Center of Gravity of a Plane Area. 

124. General differential Expressions for the Co-ordinates of a Surface of Revolution. 

125. General differential Expressions for the Co-ordinates of a Solid of Revolution. 

126. Determination of a Surface of Revolution. 

127. Determination of a Solid of Revolution. 

128. Examples — 1. Of a Circular Arc — 2. Circular Segment — 3. Surface of a Spherical 

Segment — 4. Spherical Segment. 

129. Examples on the preceding Chaptei's. 

CHAPTER VI. 

THE MECHANICAL POWERS 78 

130. Classification of the Mechanical Powers. 

§ I. THE LEVER. 

131. Definition of a Lever. 

132. Kinds of Lever. 

133. Conditions of Equilibrium when the Forces are Parallel 

134. Conditions of Equilibrium when the Forces are Inclined. 

135. Conditions of Equilibrium when the Lever is Bent or Curved. 

136. Conditions of Equilibrium when any Number of Forces in the same Plane act on 

a Lever of any Form — Examples. 

§11. WHEEL AND AXLE 86 

137. Definition of Wheel and Axle. 

138. Conditions of Equilibrium when two Forces act Tangentially to the Surface of the 

Wheel and Axle. 

139. Perpetual Lever. 

140. Pressure on the Axis. 

141. Conditions of Equilibrium of any Number of Forces. 

142. Definition of Cogged Wheels, Crown Wheels, Beveled Wheels, Pinions, Leaves. 

143. Conditions of Equilibrium in Cogged Wheels. 

144. Conditions of Equilibrium when the Cogs are of equal Breadth. 

145. Conditions of Equilibrium in Cogged Wheels and Pinions. 

§111. the cord 91 

146. Definition of the Cord — Of Tension. 

147. Conditions of Equilibrium when there are three Forces. 
1-48. Conditions of Equilibrium when the Ends are fixed. 

149. Conditions of E quilibrium when the Ends are fixed and a third Force applied to 

a Running Knot. 

150. Conditions of Equilibrium when there is any number of Forces. 

151. Relations of the Forces when in Equilibrium. 

152. Relations of the Forces when the Ends are Fixed and the Forces Parallel. 

153. Relations of the Forces when the Ends are Fixed and the Forces are Weights. 

154. Point of Application of the Resultant. 

155. Catenary — Examples. 

§ IV. the pulley 96 

156. Definition of the Pulley — Fixed and Movable. 

157. Use of Fixed Pulley. 

158. Equilibrium in single Movable Pulley. 

159. Systems of Pulleys. 

160. Equilibrium in first System. 

161. Equilibrium in the second. 

162. Equilibrium in the third — Examples. 



CONTENTS. IX 

Art Page 

§ V. THE INCLINED PLANE .. 100 

163. Definition of Inclined Plane. 

164. Equilibrium when the Body is sustained by a Force acting in any Direction. 

165. Equilibrium when the Body is sustained by a Force parallel to the Plane. 

166. Equilibrium when the Body is sustained by a Force parallel to the Base of the Plane. 

167. Equilibrium when two Bodies rest on two Inclined Planes. 

§ VI. THE WEDGE 103 

168. Definition of the Wedge — Faces, Angle Back. 

169. Conditions of Equilibrium in the Wedge. 

170. Defect in the Theory. 

171. Illustrative Problem. i 

$ VII. the screw 105 

172. Definition of the Screw. 

173. Conditions of Equilibrium in the Screw. 

$ VIII. BALANCES AND COMBINATIONS OF THE MECHANICAL POWERS 107 

174. The common Balance. 

175. Requisites for good Balance. 

176. Conditions of Horizontality of the Beam. 

177. Conditions of Sensibility. 

178. Conditions of Stability. 

179. Relations of the Requisites. 

180. Steelyard Balance. 

181. Law of Graduation. 

182. Bent Lever Balance. 

183. Law of Graduation. 

184. Roberval's Balance. 

185. Condition of Equilibrium. 

186. Conditions of Equilibrium in a Combination of Levers. 

187. Conditions of Equilibrium in the Endless Screw. 

188. Conditions of Equilibrium in any Combination of the Mechanical Powers. 

189. Conditions of Equilibrium in the Knee. 

CHAPTER VII. 

APPLICATION OF THE PRINCIPLE OF VIRTUAL VELOCITIES TO THE MECHANICAL 
POWERS 116 

190. Preliminary Considerations. 

191. Application to the Wheel and Axle. 

192. Application to Toothed Wheels. 

193. Application to Movable Pulley with Parallel Cords. 

194. Application to the first System of Pulleys. 

195. Application to the second System. 

196. Application to the third System. 

197. Application to the Inclined Plane. 

198. Application to the Wedge. 

199. Application to the Lever of any Form. 

200. Application to the single Movable Pulley with inclined Cords. 

CHAPTER VIII. 

FRICTION .124 

201. Definition of Friction— Kinds. 

202. Measurement of Friction. 

203. Laws of Friction. 

204. Value of the Coefficient of Friction. 



X CONTENTS. 

Art. Page 

205. Limits of the Ratio of the Power to the Weight on the Inclined Plane. 

206. Limits of the Ratio of the Power to the Weight in the Screw. 

207. One Limit obtained directly from the other. — Examples. 
203. Examples on Chapters VI., VII., and VIII. 



DYNAMICS. 

INTRODUCTION 131 

209. In Dynamics, Time an Element. 

210. Definition of Motion. 

211. Definition of Absolute Motion. 

212. Definition of Relative Motion. 

213. Definition of Velocity — Its Measure. 

214. Definition of Variable Velocity — Its Measure. 

215. Definition of Relative Velocity. 

216. Definition of Inertia — First Law of Motion. 

217. Definition of Center of Inertia. 

218. Definition of the Path of a Body. 

219. Definition of Free and Constrained Motion. 

220. Definition of an Impulsive Force. 

221. Definition of an Incessant Force. 

222. Definition of a Constant Force — Its Measure. 

223. Definition of a Variable Force — Its Measure. 

224. Definition of Momentum — Its Measure — Of Living Force. 

225. Definition of a Moving Force — Its Measure. 

226. The secona Law of Motion. 

227. The third Law of Motion. 

CHAPTER I. 

UNIFORM MOTION 136 

228. Point tc which the Force must be applied. 

229. General Equation of Uniform Motion. 

230. Relation of Spaces to Velocities when the Times are Equal. 

231. Relation of Spaces to the Times when the Velocities are Equal. 

232. Relation of Velocities to the Times when the Spaces are Equal. 

233. Measure of an Impulsive Force. 

234. The Velocity resulting from the Action of several Forces. 

235. Parallelogram, cf Velocities. 

236. Rectangular Composition and Resolution of Velocities. 

237. Relations of Space, Time, and Velocity of two Bodies moving in the same Straight 

Line. 

238. Relations of Space, Time, and Velocity of two Bodies moving in the Circumference 

of a Circle. 

239. Examples. 

CHAPTER II. 

IMPACT OF BODIES.... ................ 143 

240. Definition of Direct, Central, and Oblique Impact. 

241. Definition of Elasticity — Perfect — Imperfect — Its Modulus. 

242. Definition of Hard and Soft. 

243. Velocity of two Inelastic Bodies after Impact. 

244. Loss of Living Force in the Impact of Inelastic B 

245. Velocities of imperfectly Elastic Bodies after Im; . . 



CONTENTS. XI 

Art. Pa§e 

246. Velocity of the nth Body in a Series of perfectly Elastic Bodies. 

247. Velocity of the common Center of Gravity before and after Impact. 

248. Conservation of the Motion of the Center of Gravity. 

249. Definition of Angles of Incidence and Reflection. 

250. Motion of an Inelastic Body after Oblique Impact on a Hard Plane. 

251. Motion of an Elastic Body after Oblique Impact on a Hard Plane. 

252. Direction of Motion before Impact, that a Body after Impact may pass through a 

given Point. 

253. Measure of the Modulus of Elasticity. 

254. Mode of detemnning it — Table of Moduli. 

255. Examples. 

CHAPTER III. 

MOTION FROM THE ACTION OF A CONSTANT FORCE 153 

256. Uniformly accelerated Motion — Acquired Velocity. 

257. Space in Terms of the Force and Time. 

258. Space in Terms of the Force and Velocity. 

259. Space described in the last n Seconds. 

260. The Velocity and Space from the joint Action of a Projectile and Constant Force. 

261. The Velocity when the Space is given. 

262. Velocity lost and gained by the Action of a Constant Force when the Space is 

the same. 

263. Scholium on Universal Gravitation. 

264. Scholium on the Numerical Value of the Force of Gravity. 

265. Examples. 

CHAPTER IV. 

P R OJ E CT I L E 3 161 

266. The Path of a Projectile is a Parabola. 

267. Equation of the Path when referred to Horizontal and Vertical Axes. 

268. Definition of Horizontal Range — Time of Flight — Impetus. 

269. Time of Flight on a Horizontal Plane. 

270. Range on a Horizontal Plane — The same for two Angles of Elevation. 

271. Greatest Height. 

272. Range and Time of Flight on an Inclined Plane, and Co-ordinates of Point of Impact. 

273. Formula for Velocity of a Ball or Shell. 

274. Examples. 

CHAPTER V. 

CONSTRAINED MOTION. 
§ I. MOTION ON INCLINED PLANES 169 

275. Relations of Space, Time, and Velocity. 

276. Velocity down the Plane and its Height. 

277. Times down Inclined Planes of the same Height. 

278. Relations of Space, Time, and Velocity when projected up or down the Plane. 

279. Time of Descent down the Chords of a Circle. 

280. Straight Line of quickest Descent from a Point within a Circle to its Circumference. 

281. Straight Line of quickest Descent from a given Point to an Inclined Plane. 

282. Motion of two Bodies suspended by a Cord over a Fixed Pulley. 

283. Motion of two Bodies when the Inertia of the Pulley is considered. 

§11. MOTION IN CIRCULAR ARCS 173 

284. Velocity acquired down the Arc of a Circle. 

285. Velocity lost in passing from one Side of a Polygon to the next. 



Xll CONTENTS. 

Art. Page 

286. Velocity lost when the Sides are Infinite in Number. 

287. Direction and Intensity of an Impulse at each Angle, to make a Body describe a 

Polygon with a uniform Velocity. 

288. Direction and Intensity when the Polygon becomes a Circle. 

289. Definition of Centrifugal and Centripetal Force. 

290. Discussion of the Motion of a Body in a Circle by the Action of a Central Force. 

291. Centrifugal Force compared with Gravity. 

292. Centrifugal Force of the Earth at the Equator. 

293. Ratio of Centrifugal Force of the Earth to Gravity. 

294. Time of its Rotation when they are Equal. 

295. Diminution of Gravity in different Latitudes by the Centrifugal Force. 

296. Cause and Value of the Compression of the Earth. 

297. Centrifugal Force of the Moon in its Orbit. 

298. Moon retained in its Orbit by Gravitation. 

§ III. pendulum 182 

299. Definition of a Simple Pendulum. 

300. The Force by which it is urged in its Path. 

301. Times of Descent to the Center of Force equal when the Force varies as the Dis- 

tance. 

302. Expression for the Time given. 

303. Discussion of the Vibrations of Pendulums. 

304. Pendulum used to determine the Figure of the Earth. 

305. Lengths of Pendulums as the Squares of the Number of Vibrations in a given Time. 

306. Mode of determining the Length of the Seconds Pendulum. 

307. Determination of the Intensity of Gravity. 

308. Correction of the Length of the Pendulum for a given Loss or Gain. 

309. Rate at a given Height above the Earth's Surface. 

310. Heights determined by the Change of Rate. 

311. Definition of Conical Pendulum. 

312. Tension, Velocity, and Time determined. 

313. Formula for Length of Pendulum in a given Latitude. 

314. Examples. 

CHAPTER VI. 

ROTATION OF RIGID BODIES 191 

315. Angular Velocity of a Body about a Fixed Axis. 

316. Definition of Moment of Inertia. 

317. Moment of Inertia about any Axis compared with that about a Parallel Axis 

through the Center of Gravity. 

318. Radius of Gyration — Principal Radius. 

319. Existence of a Point at which, if all the Matter of the Body were collected, the 

Motion would be the same. 

320. Expression for its Distance from the Axis. 

321. Definition of Center of Oscillation. 

322. Time of Oscillation of a Rigid Body. 

323. Definition of Compound Pendulum. 

324. Centers of Oscillation and Suspension convertible. 

325. Length of an Equivalent Simple Pendulum. 

326. Value of Principal Radius of Gyration. 

327. Relations of the Centers of Gravity, of Gyration, and of Oscillation. 

328. Determination of the Length of the Seconds Pendulum. * 

329. Examples on the Relation of the Simple to the Compound Pendulum. 

330. Examples on the Determination of the Moment of Inertia. 



CONTENTS. Xlli 



Paga 

CHAPTER VII 201 



331. More General Methods required. 

332. Modification and Extension of Fundamental Formulae. 

§ I. RECTILINEAR MOTION OF A FREE POINT 202 

333. Preliminary Considerations — Absolute Force. 

334. Space in Terms of the Time when the Force is Constant. 

335. General Expression for the Velocity when the Force is Variable. 

336. General Expression for the Time when the Force is Variable. 

337. Velocity when the Force varies as the Distance. 

338. Velocity at the Center of the Earth. 

339. Time when the Force varies as the Distance. 

340. Velocity when the Force varies as the Square of the Distance. 

341. Velocity from an Infinite Distance. 

342. Time when the Force varies inversely as the Square of the Distance. 

343. Velocity and Time when the Force varies inversely as the Cube of the Distance. 

§ II. CURVILINEAR MOTION OF A FREE POINT 209 

344. Preliminary Considerations. 

345. Motion of a Point acted on by any Number of Forces in the same Plane. 

346. Specific Directions for particular Cases. 

347. Velocity of a Point in its Path. 

348. The Velocity may be found when the Expression is integrable. 

349. The Expression integrable when the Forces are directed to Fixed Centers, and 

are Functions of the Distances from those Centers. 

350. Motion of a Point when the Force is directed to a Center. 

351. When there is an Equable Desci'iption of Areas, the Force is directed to a Center. 

352. Motion of a Point from an Impulsive Force. 

353. Motion of a Projectile acted upon by Gravity. 

§ III. CONSTRAINED MOTION OF A POINT 215 

354. Velocity of a Point on a given Curve. 

355. Time of its Motion. 

356. The Reaction of the Curve. 

357. Motion down the Arc of a Circle by Gravity. 

358. Motion down the Arc of a Cycloid. 

359. Tautochronism of Vibrations in a Cycloid. 

§ IV. MOTION OF A POINT ACTED ON BY A CENTRAL FORCE 221 

360. Preliminary Remarks. 

361. Expressions for the Force in Terms of Polar Co-ordinates. 

362. Equable Description of Areas. 

363. Angular Velocity of the Radius Vector. 

364. Velocity of a Point in its Orbit. 

365. Time of describing any Portion of its Orbit. 

366. Equation of the Orbit. 

367. Geometrical Values of Differential Expressions. 

368. Kepler's Laws. 

369. Accelerating Force of a Planet directed to the Sun. 

370. Accelerating Force of a Planet varies inversely as the Square of the Distance. 

371. Absolute Force the same for all the Planets. 

372. Velocity of a Planet at any Point of its Orbit. 

373. Determination of the Orbit when the Force varies inversely as the Square of the 

Distance. 

374. Time of describing any Portion of the Orbit. 



XIV CONTENTS. 

Art. Page 

375. Value of the Absolute Force in Terms of the Masses— In Terms of Terrestrial 

Gravity. 

376. Examples. 



HYDROSTATICS. 

CHAPTER I 236 

377. Forms of Matter. 

378. Definition of a Perfect Fluid. 

379. Definition of an Incompressible Fluid. 

380. Definition of a Compressible Fluid. 

381. Unimpaired Transmission of Pressures. 

382. Surface of a Fluid at Rest. 

383. Surface of a Fluid moving Horizontally. 

384. Surface of a Fluid Revolving. 

385. Pressure on tbe Horizontal Base of a Vessel. 

386. Pressure on any Side of a Vessel or immersed Surface. 

387. Pressure in a definite Direction. 

388. Resultant Pressure on tbe Interior Surface of a Vessel. 

389. Resultant Pressure on an immersed Solid. 

390. Conditions of Equilibrium of an immersed Solid. 

391. Definition of Plane of Flotation — Of Axis of Flotation. 

392. Depth of Flotation of a Body. 

393. Position of Equilibrium of a Triangular Prism. 

394. Definition of Center of Pressure. 

395. Formula for finding the Center of Pressure. 

396. Center of Pressure of a Rectangle with one Side in the Surface of the Fluid. 

397. Center of Pressure of a Triangle with tbe Vertex in the Surface. 

398. Center of Pressure of a Triangle with a Side in the Surface. 

399. Center of Pressure of a Rectangle wholly immersed. 

400. Equilibrium of Fluids of different Density in the same Vessel. 

401. Equilibrium of Fluids of different Density in a bent Tube. 

402. Examples. 

CHAPTER II. 

SPECIFIC GRAVITY 264 

403. Definition of Specific Gravity. 

404. Specific Gravity of a Body more Dense than Water. 

405. Specific Gravity of a Body less Dense than Water. 

406. Specific Gravity of a Liquid. 

407. Weights of the Constituents of a Mechanical Mixture. 

408. Hydrometer. 1°. Of Constant Weight. 2°. Of Constant Volume. 

409. Nicholson's Hydrometer. 

410. Examples. 

CHAPTER III. 

COMPRESSIBLE OR AERIFORM FLUIDS 272 

411. Tension of a Compressible Fluid found. 

412. The Unit of Pressure. 

413. Tension inversely as the Volume. 

414. Tension directly as the Density. 



CONTENTS. XV 

Art. Page 

415. Effect of Heat on "Volume and Tension. 

415. Relation of Density to Temperature and Pressure. 

416. Density of the Atmosphere in Terms of the Height. 

417. Barometrical Measurement of Heights. 

418. Examples. 



HYDRODYNAMICS. 

419. Velocity of a Fluid in a Tube of variable Diameter 281 

420. Velocity of a Fluid from a small Orifice in the Bottom of a Vessel. 

421. Horizontal Range of a Spouting Fluid. 

422. Quantity of Discharge from a small Orifice. 

423. Time required for a Vessel to empty itself. 

424. Vena Contracta — Coefficient of Efflux. 

425. Discharge from a Rectangular Aperture. 

426. Discharge from a Triangular Aperture. 

427. Velocity of Efliux of an Elastic Fluid. 

428. Motion of Fluids in Long Pipes. 

429. General Method of determining the Discharge from small Orifices. 

430. General Method of determining the Time for a Vessel to empty itself. 

431. General Method of determining the Discharge from Orifices of any Form. 

432. Examples. 



HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 

433. Mariotte's Flask 295 

434. Bramah's Press. 

435. Hydrostatic Bellows. 

436. Diving BelL 

437. Sea Gage. 

438. Siphon. 

439. Common Pump. 

440. Common Pump, Conditions of Failure. 

441. Air Pump. 

442. Condense!*. 

443. Clepsydra. 



ELEMENTARY MECHANICS, 



INTRODUCTION. 

1. Mechanics is the Science which treats of the laws of 
Equilibrium and Motion. It is subdivided into Statics, Dy- 
namics, Hydrostatics, and Hydrodynamics. 

Statics treats of the necessary relations in the intensities and 
directions of forces, in order to produce equilibrium of solid 
bodies. 

Dynamics treats of the effects of forces on solid bodies when 
motion is produced. 

Hydrostatics investigates the conditions of equilibrium in 
fluid bodies. 

Hydrodynamics investigates the effects of forces on fluids 
when motion results.* 

2. Force is that which produces or tends to produce motion 
or change of motion. 

The consideration of the nature of force does not belong to 
the present subject. Mechanics is concerned only with the 
effects of force as exhibited in the production of motion or rest- 

3. The effect of a force depends on, 1st, its Intensity; 2d, its 
Direction ; and, 3d, its Point of Application. 

The Intensity of a force may be measured, statically, by the 
pressure it will produce, or by the weight which will counter- 
poise it ; dynamically, by the quantity of motion it will produce. 
By assuming for a unit of force that force which is counter- 
poised by a known weight, the intensity or magnitude of any 
other force will be expressed by the numerical ratio which its 
counterpoise will bear to the counterpoise of the unit of force. 

* Theoretic Statics and Dynamics are those branches of theoretic Mechanics which 
treat of the effects of forces applied to material points or particles regarded as without 
weight or magnitude. Static and Dynamic Somatology would then embrace the appli- 
cation of theoretic Statics and Dynamics to bodies of definite form and magnitude, both 
solid and fluid. 



2 INTRODUCTION. 

In the same manner, by fixing on any line to represent the unit 
of force, any other force will be represented by the line which 
bears to the linear unit the same ratio which the force in 
question bears to the unit of force. 

The Direction of a force is the line which a material point, 
acted on by that force, would describe were it perfectly free. 

The point of application of a force is that point in its line of 
direction on which the force acts. 

4. As the magnitudes, directions, and points of lines are all 
determinable by the principles of Analytical Geometry, so 
forces, of which lines are the appropriate representatives, 
come under the dominion of the same principles. The appli- 
cation of these principles to the determination of the laws of 
equilibrium and motion, considered as the effects of forces, 
constitutes analytical mechanics. 

5. Forces may act on a point either by pushing or pulling it. 
As these are readily convertible, the one into the other, with- 
out affecting the intensity, direction, or point of application of 
the forces themselves, all forces will be regarded as pulling 
unless otherwise expressly stated. 

For convenience, we shall call those forces which have a 
common point of application concurring forces, and those 
which act along the same line toward the same parts, con- 
spiring forces. 

6. A body is an assemblage of material points. The material 
points, or elementary particles, are connected together in vari- 
ous, ways, according to the nature of the body. 

A body is said to be rigid when the relative position of its 
particles remains unchanged by the action of forces upon it. 

In flexible and elastic bodies the relative positions of the 
particles change by the action of forces. 



STATICS 



CHAPTER I. 

ON THE COMPOSITION AND EGIUILIBRIUM OF CONCURRING FORCES, 

7. Prop. Two equal forces applied to the same point in ex- 
actly opposite directions are in equilibrium. 

For no reason can be assigned why motion should take 
place in the direction of one force which will not equally ap- 
ply to the other. 

8. Prop. Two concurring forces forming an angle with each 
other can not be in equilibrium. 

If possible, let the two forces P and Q, acting at A, and 
making, by their directions, the an- p^ 
gle PAQ, be in equilibrium. Apply 

to the point A the force P , , equal and -< ^- > 

opposite to Q. Since, by hypothe- Pa Q 

sis, the forces P and Q, are in equilibrium, the point A will 
move in the direction of P,. But P,"and Q, being equal and 
opposite {Art. 7), will be in equilibrium, and hence the point 
A must move in the direction of P. Therefore the point A 
must move in two directions at once, which is absurd. 

9. Definition. The resultant of two or more forces is a 
force which singly will produce the same mechanical effect as 
the forces themselves jointly. 

The original forces are called components. 
Cor. In all statical investigations the components may be 
replaced by their resultant, and vice versa. 

10. Prop. The resultant of several conspiring forces is a 
single force. equal to their sum, and acting in the same direction. 



STATICS. 



This is obvious from the admitted mode of measuring forces. 

1 1 . Prop. The resultant of two unequal forces acting in op- 
posite directions is a single force equal to their difference, and 
acting in the direction of the larger. 

For the smaller will obviously be expended in annulling in 
the larger a quantity equal to itself, and thus leave an effective 
balance equal to the excess of the larger over the smaller, and 
acting in the direction of the larger. 

12. Prop. The resultant of any number of forces acting in 
the same right line is equal to their algebraic sum. 

By Art. 10, the resultant of each system of conspiring for- 
ces is equal to their sum, and by Art. 11, the resultant of 
these two resultants is equal to their difference with the sign 
of the greater prefixed. 

13. Prop. In a system of points invariably connected, any 
point in the direction of a force may be taken as the point of 
application. 

If the force P be applied at the point B, and to any othei 
» -d p point A in its direction two 

j j »■ other forces be applied, 

each equal to it, but opposite to each other, since B, by hypoth- 
esis, is invariably connected with A, one of these forces {Art. 
7) will be in equilibrium with P. There will remain, there- 
fore, a single effective force equal to P and applied at A. In 
the same manner it may be transferred to any other point in 
its direction. 

14. Prop. The resultant of several concurring forces in one 
plane lies in the same plane. 

For if we suppose the resultant to lie out of the plane of the 
forces on one side, we may always conceive a line symmet- 
rically situated on the other side of the plane ; and since no 
reason can be assigned why it should be in one of these lines 
rather than in the other, it can be in neither of them, unless we 
admit the absurd consequence, that it is in both ; in other 
words, that a system of forces has two resultants. 

15. Prop. The resultant of two equal concurring forces is in 



CONCURRING FORCES. 5 

the direction of a line bisecting the angle formed by the compo- 
nents. 

For no reason can be assigned why it should tend to one 
side rather than the other of this line. 

16. Cor. When the forces are unequal, it is obvious that the 
direction of the resultant will make a less angle with the larger 
force than with the smaller, and the greater the disparity in 
the forces, the smaller will be this angle. 

17. Prop. If all the forces of a system, while their directions 
are preserved, are increased or diminished in any ratio, their 
resultant, without changing its direction, will be varied in the 
same ratio ; and if the components were previously in equilib- 
rium, they will remain so in whatever ratio their intensities be 
varied. 

Let Pj, P 2 , P 3 be any concurring forces whatever. 

Since (Cor., Art. 9) we can replace them by their resultant 
R, if we double or treble each of the components, this will only 
double or treble the resultant R, without changing its direction, 
because, in so doing, we only add to the system one or more 
systems precisely equal to the first. 

In like manner, if we reduce the original components to one 
half or one third of their former intensities, the resultant will 
still preserve its direction, but will become one half or one 
third as large as before ; since to double or treble these is only 
to double or treble their resultant, and thus reproduce the 
original system. 

If the forces are in equilibrium, by varying the magnitudes 
of all in the same ratio, we only add to or suppress from the 
system other systems already in equilibrium. 

18. Prop. Three equal concurring forces, inclined at angles 
of 120°, will be in equilibrium. 

Since each force is inclined in the same Pl ^/ 

angle to the directions of the other two, 
any reason that can be assigned why ei- 
ther one will prevail, will apply with 
equal force to show that each of the oth- 
ers will prevail. ^3 





6 STATICS. 

19. Prop. Two equal concurring forces, inclined at an angle 
of 120°, have for their resultant a force which will be repre- 
sented in magnitude and direction by the diagonal of a rhombus 
constructed on the lines representing the components. 

Let AD and AB represent the two forces T? 1 and P 2 , mak- 
ing the angle BAD=120°. To the 
D% point A apply a force P 3 = P, or P 2 , 

and making, with P, and P 2 , angles of 
120°. By Art. 18, these three forces 
R will be in equilibrium. Now since P x 
and P 2 equilibrate P 3 , their resultant 
must also. Therefore produce FA to 
C, making AC=AF; and since AC=R 
equilibrates P 3 , it will be the resultant of P, and P 2 . Join 
DC and BC, and ABCD will be a rhombus, of which AC is 
the diagonal. For since CAD— 60°, and AC=AD, the triangle 
CAD is isosceles ; and since each angle is equal to 60°. it is 
equilateral. .*. DC = AD=AB. In the same manner it may be 
shown that CB=AD, and hence the figure is a rhombus. 

Cor. Denoting the angle BAC by a, we have AE=AB 
cos. a—V cos. a, and the resultant AC=R=2P cos. a. 

20. Prop. If three forces are in equilibrium, each will be equal 
and opposite in direction to the resultant of the other two. 

For, replacing either two by their resultant, we shall have 
two forces in equilibrium. If they be not opposite in direc- 
tion, by Art. 8, they can not be in equilibrium ; and if they be 
not equal, by Art. 11, motion will ensue. 

PARALLELOGRAM OF FORCES. 

21. Prop. If two forces be represented in magnitude and di- 
rection by the two adjacent sides of a parallelogram, the diago- 
nal will represent their resultant in magnitude and direction. 

First. The direction of the diagonal is that of the resultant. 

1°. When the forces are equal, this is obvious from Art. 15. 

2°. Let us assume, for the present, that it is also true for the 
two systems of unequal forces P and Q,, and P and R ; then it 
will be true for the forces P and Q+R. Let P and Q act 




CONCURRING FORCES. 1 

at A in the directions AD and AB respectively, and be repre- 
sented in magnitude by these lines, a Q. Q3 R . 
Suppose the force R to act at B 
(Art. 13), a point in its direction, 
and to be represented in magnitude 
by BC. Complete the parallelo- D - E F 

gram ADFC, draw BE parallel to AD, and the diagonals AE, 
AF, and BF. Then P at A, in AD, and Q, at A, in AB, have, 
by hypothesis, their resultant in the direction AE, and (Art. 
13) may be supposed to act at E. Replacing this resultant by 
its components acting in their original directions, we have a 
force P acting at E in the direction BE, and a force Q acting 
at E in the direction EF. Transfer P to B and Q to F, 
without changing their directions. Then P at B, in BE, and 
R at B, in BC, will also, by hypothesis, have their resultant in 
the direction BF, which may be supposed to act at F. We 
now have all the forces acting at F, and this without disturb- 
ing their effect upon the point A, supposed to be invariably 
connected with F. Hence, if the assumption be correct, F is 
a point in the direction of the resultant of the forces P and Q, 
+R. But the assumption is correct when Q and R are each 
equal to P (Art. 15). Therefore, the proposition for the direc- 
tion of the resultant is true for P and 2P. Again, making Q, 
=2P and R=P, it is true for P and 3P, and so for P and nP. 
Also, putting nY for P, Q=P, and R=P, it will be true for tzP 
and 2P, and so on for mY and ?iP (m and n being positive in- 
tegers), or for all commensurable forces. 

3°. When the forces are incommensurable. Let AB and 

AC represent the forces. Complete A ^b 

the parallelogram, and draw the diag- 
onal AD. AD will represent the di- 
rection of the resultant. If not, let 
some other line, as AE, be its direc- 
tion. Divide AB into a number of 
equal parts less than DE, and on AC c~ 

take as many of these parts as possible. Since AC and AB 
are incommensurable, there will be a remainder GC less than 
DE. Draw GF parallel to AB, and join AF. AB and AG 





8 STATICS. 

representing two commensurable forces, AF will represent the 
direction of their resultant. But (Art. 16) the resultant of AB 
and AC will make a less angle with AC than the resultant of 
AB and AG does. .*. AE can not be the direction of the re- 
sultant of AB and AC ; and similarly, it can be shown that no 
other direction than AD can be that of the resultant. 

Second. The diagonal will represent the magnitude of the 
D resultant. Let P, Q, and R 

be three forces in equilibrium, 
and AD, AB, AF represent 
their magnitudes respective- 
ly. Complete the parallelo- 
grams BD and DF, and draw the diagonals AC and AE. AC 
being the direction of the resultant of P and Q,, must, by Art 
20, be in the same straight line with AF, and is therefore par- 
allel to DE. In the same manner, AE being the direction of 
the resultant of P and R, will be in the direction of BA pro- 
duced, and therefore parallel to CD. Hence CAED is a par- 
allelogram, and CA is equal to DE, which is equal to AF, by 
construction. But, by Art. 20, AF represents the magnitude 
of the resultant of P and Q. Hence CA, the diagonal of the 
parallelogram constructed on the lines AD and AB, represents 
the resultant of P and Q, in magnitude as well as direction. 

TRIANGLE OF FORCES. 

22. Prop. If three concurring forces are in equilibrium, and 
a triangle be formed by lines drawn in their directions, the sides 
of the triangle, taken in order, will represent the forces. Con- 
versely, if the forces can be represented by the sides of a triangle, 
taken in order, they will be in equilibrium. 

Let the forces P, Q, and R be in equilibrium, and be repre- 
sented by AB, AD, and AE respectively. 
Produce EA, draw BC parallel to AD, 
, and complete the parallelogram. AC 
will represent the resultant of P and Q 
(Art. 21) ; and since R equilibrates P 
and Q, it must be equal and opposite to 
EA=AC. And since BC is equal, and 



E 


R. A/ \ 


^... 


V 


D 

their resultant. .*. 



CONCURRING FORCES. 9 

also parallel to AD, that is, has the same direction, the three 
sides of the triangle ABC taken in order represent the three 
forces P, Q, and R. 

Conversely. If the three sides AB, BC, and CA of the tri- 
angle ABC, taken in order, represent the direction and magni- 
tude of the three forces P, Q,, and R, they will be in equilibri- 
um. Draw from A a line parallel to BC, and from C a line 
parallel to AB, meeting the former in D. The resultant of 
the two forces P and Q, represented by AB and BC, or AB 
and AD, is equal to, and in the direction of the diagonal AC ; 
that is, equal and opposite in direction to the force R. Hence 
P, Q, and R are in equilibrium. 

23. Cor. Hence a given force may be resolved into two 
component forces, acting in given directions. Also into two 
others, one of which is given in magnitude and direction. 

1°. Let AB be the given force, and 
AC, AD the given directions ; that is, 
making known angles with AB. From 
B draw BF, and BE parallel to AD and 
AC. Then AF and AE are the two A 
components (Art. 21) acting in the given 
directions AC and AD. ** "^D 

2°. To resolve AB into two others, one of which shall be in 
the direction AC, and be equal to AF. Draw FB, and com- 
plete the parallelogram. AE will be the other component. 

When the directions of the components are arbitrary, their 
valuation will be most easily effected by assuming these di- 
rections at right angles. If the angle CAD be right, then AE 
=AB cos. BAE, and AF=AB cos. BAF. 

POLYGON OF FORCES. 

24. Prop. If any number of concurring forces be represented 
in magnitude and direction by the sides of a polygon, taken in 
order, they will be in equilibrium. 

Let the sides of the polygon ABCDEx\ represent the mag- 
nitudes and have the directions respectively of the forces P a , 
P 2 , P 3 , P 4 , and P 5 . These forces will be in equilibrium. 




10 



T A TICS. 




Draw the diagonals AC and AD. By Art. 22, AC, the third 

side of the triangle ABC, repre- 
sents a force equivalent to the 
two forces AB and BC. AC is 
therefore the resultant of P x and 
P 2 , and may be substituted for 
them. In the same manner, AD 
is equivalent to the forces AC 
J I> and CD, or to P lf P 2 , and P 3 ; 
and AE to the forces AD and 
DE, or to the forces P,, P 2 , P 3 , 
and P 4 . That is, AE is the re- 
sultant of the four forces P,, P 2 , P 3 , and P 4 , and acts in the 
direction AE. But P 5 is represented in magnitude by EA, and 
acts in the direction EA. Hence P 5 is equal and opposite to 
the resultant of the other four, or the forces are in equilibrium. 
Coit. The proposition is true of forces which do not lie all in 
one plane. For the proof is independent of this supposition. 

25. Scholium. When three forces are in equilibrium, any 
three lines, taken parallel to their directions, will form a trian- 
gle, the sides of which respectively will represent the relative 
magnitudes of the forces ; but when there are four or more 
forces this will not hold, since the relation which subsists be- 
tween the sides and angles of triangles does not obtain in poly- 
gons of more than three sides. 

26. Prop. To find, graphically, the resultant of any number 
of forces, acting at different points in the same plane. 

Let the forces P 1? P 2 , P 3 have their points of application at 
A, B, and C. Producing the directions of 
the forces P x and P 2 till they meet in D, 
construct the parallelogram AB on the 
lines representing them, and the diagonal 
will represent their resultant R^ Pro- 
ducing R , until it meets the direction of 
P 3 in F, and constructing the parallelo- 
gram on the lines representing R } and P 3 , 
the diagonal will represent their resultant 




CONCURRING FORCES. 11 

R 2 , or the resultant of P 15 P 2 , and P 3 . By the same process 
we may find the resultant of any number of forces. If any 
two are parallel, we must first compound one of these with a 
third, whose direction is inclined to it. If all are parallel, the 
case will belong to parallel forces. 

PARALLELOPIPED OF FORCES. 

27. Prop. If three concurring forces lying in different planes 
he represented in magnitude and direction by the three edges of 
a parallelopiped, then the diagonal will represent their resultant 
in magnitude and direction ; and conversely, if the diagonal 
represents a force, it is equivalent to three forces represented by 
the edges of the parallelopiped. 

Let the three edges AB, AC, AD of the parallelopiped rep- 
resent the three forces. Then AE, B 
the diagonal of the face ACED, 
represents the resultant of the 
forces AD and AC. Compound- 
ing this with the third force, rep- 
resented by AB, we have AF, the c~ 
diagonal of the parallelogram AEFB, for the resultant of AE 
and AB, or of the forces AC, AD, AB. 

Reciprocally, the force AF is equivalent to the components 
AB, AE, or to the components AB, AC, and AD. 

28. Prop. If three forces are in equilibrium, they are propor- 
tional each to the sine of the angle made by the directions of the 
other two. 

By Art. 22 and figure, 

P : Q : R=AB : BC 

and by trig., =sin. BCA 

= sin. CAD 

or because sin. A= sin. (180° — A), 

= sin. DAE 

=sin. Q.R 

29. Prop. If P and Q, be two concurring forces, 6 the angle 
made by their directions, and R their resultant, then R 2 =P 2 -f 
Q 2 +2PQ cos. 6. 




:CA 


"» 






: sin. 


CAB: 


sin. 


ABC; 


: sin. 


CAB: 


i sin. 


ABC; 


: sin. 


EAB: 


sin. 


BAD; 


: sin. 


PR : 


sin. 


PQ. 



12 



STATICS. 



By trig, we have, from Jig., Art. 22, 

AC 2 =AB 2 +BC 2 -2AB.BC cos. ABC, 
and cos. ABC= — cos. BAD= — cos. 0. 

R 2 =P 2 +Q 2 +2PQ cos. 6. 

30. Def. The moment of a force about any point is the prod- 
uct of the force into the perpendicular let fall from that point 
on the direction of the force. The point is called the origin 
of moments. 

The moment of a force measures the tendency of the force 
to produce rotatory motion about a fixed point. 

31. Prop. The moment of the resultant of two forces equals 
the algebraic sum of the moments of the components. 

1°. When the origin of moments falls without the angle 
made by the forces. 

Let AB, AC represent the two 
forces P and Q, and AD the diago- 
nal of the parallelogram constructed 
on AB and AC, their resultant R. 
H Take any point, O, for the origin of 
moments ; join OA and draw AG per- 
pendicular to OA ; draw 01, Om, On 
respectively perpendicular to AB, AC, 
AD, and BE, CF, DG perpendicular 
to AG ; also CH parallel to AG. 

The triangles OZA, OmA, OnA are respectively similar to 
the triangles AEB, AFC, AGD. 

Whence AE : AB =0/ : OA, or AE=^^, 




G 



AF : AC -Om : OA, or AF: 



AG : AD=0?i : OA, or AG: 



OA 
AC.Om 
"OX"' 
AP.Otz 

OA " 



But the triangles ABE and CDH being equal, AE=CH=FG. 
.-. AE+AF=AG, 
or AB.OZ+AC.Om=AD.07i, or designating 01, Om, and On by 
p, q, and r respectively, V.p+Q.q=R.r. 

2°. When the origin of moments is taken within the angle. 



CONCURRING FORCES. 



13 



1/ 




ow- 











V?7 




>JQ 











In this case the moments of the forces tend to produce rota 
tion in opposite directions. As- ^ A G ¥ 

suming one direction as positive, 
to distinguish them we must regard 
the other as negative. Let the 
positive direction be that of the 
hands of a watch ; then Q,.0??i will 
be positive, and ¥.01 negative. 

The proof is the same in this 
case, except that AG=AF— AE, 

and AC.Om-AB.OZ=AD.07i, 

or Q.9— P./?=R.r. 

32. Cor. 1. The moment of the resultant of any number of 
concurring forces in the same plane, is equal to the algebraic sum 
of the moments of the components. 

By compounding the resultant R with a third force, we 
should obtain a like result. In the same manner, the proposi- 
tion may be extended to any number of forces. 

33. Cor. 2. If the origin of moments be a fixed point, and 
taken in the direction of the resultant, On will become zero, 
and 

P.OZ=Q.O™, or Y.p=Q.q; 
that is, while the fixed point O, by its resistance, counteracts 
the resultant force, the forces P and Q, will be in equilibrium 
about that point, since their moments, tending to cause rota- 
tion in opposite directions, are equal. 

34. Cor. 3. If several forces are in equilibrium, the resultant 
force R is zero, and the moment of the resultant R.r=0. 
Hence the moments in one direction balance those in the op- 
posite direction, and there is no tendency to motion either of 
translation or rotation. 



35. EXAMPLES. 

1. When the component forces are P and Q, and the angle 
made by their directions 6 ; what is the magnitude of the re- 
sultant R when 0=0 and d=nl 

Ans. (P+Q) and (P-Q). 



14 STATICS. 

2. Show that R is greatest when 0=0, least when 6=tt, and 
intermediate for intermediate values of 0. 

3. When P=Q and (9=60°, find R. 

Arts. R=Pv'3. 

• 4. When P=Q and 0=135°, find R. 

Ans. R=P\/2-v / 2. 

* 5. When the three concurring forces 3m, 4m, and 5m are 
in equilibrium, find the angle 3m, 4m. 

Ans, 90°. 

6. If P=Q and 6=120°, find R. 

Ans. R=P. 

7. If P=6, Q=ll, and 6=30°, find the magnitude of R, and 
of the angles P,R and Q,R. 

Ans. R= 16.47, P?R=19°.30', Q?R=10°.30'. 

8. Apply the proof of the polygon of forces to the case of 
five equal forces represented by the sides of a regular penta- 
gon taken in order. 

9. A cord is tied round a pin at the fixed point A, and its 
two ends are drawn in different directions by the forces P and 
Q,. Find the angle P,Q=0, when the pressure on the pin is 

Ans. Cos. 6= -^ . 

10. A cord, whose length is 21, is tied at the points A and B 

in the same horizontal line, whose distance is 2a; a smooth 

ring upon the cord sustains a weight w : find the force T of 

tension in the cord. 

™ w 

Ans. T= 



a' 



* 11. Given the four concurring forces P^l? P 2 = 2, P 3 =3, 
P 4 =4, and the angles P7vP 3 = 90°, P^P 4 =90°, and P^P 2 = 
60°. Find the magnitude of the resultant R, and its direction 

pTX 

Ans. R=6.889 and P^R=102°.16'. 



CHAPTER II. 



PARALLEL FORCES. 

Having considered the subject of forces which have a com- 
mon point of application, or which are reducible to this condi- 
tion, we next proceed to consider forces which act on differ- 
ent points, connected together in some invariable manner, as 
in rigid bodies, and whose directions are parallel. 

36. Prop. The resultant of two parallel forces, acting at the 
extremities of a rigid right line, is parallel to the components, 
equal to their algebraic sum, and divides the right line, or the right 
line produced, into parts reciprocally proportional to the forces. 

1°. When the forces act in the same direction. 

Let the parallel forces P and V x act at A and B. At these 
points apply the equal and 
opposite forces P' and P' , ; p -<- 

these will not disturb the 
system. P and P' at A 
will have a resultant P", 
and Pj and P', at B, a re- 
sultant P",. The direc- 
tions of P" and P", will 
meet in some point D, at 
which we may suppose 
them to act. Replacing P" by P and P', and P", by P 1 and 
P'j, we now have, acting at D, the four forces P, P', P,, 
and P' l , of which P' and P' , are equal and opposite, while P 
and Pj act in the direction DC, and have a resultant R=P+ 
Pj (Art. 10). To determine the point C in the line AB, where 
R acts, since the sides of the triangle ACD have the directions 
respectively of the forces P, P', and their resultant P" (Art 
22), we have, 




E=P+p L 



16 STATICS. 

P : F=DC : CA, 

and similarly from triangle BCD, 

P, rP^BCiDC. 
Compounding these ratios, we have, since P / =P / 1 , 

P : P^BC : CA, or P.AC^.BC. 
2°. When the forces act in opposite directions. 
Let P and P 2 (Pj>P) be two parallel forces, acting at A 

and B in opposite directions, 

"P T?~"P-~P ~P" 

and apply to these points, as 
before, the equal and opposite 
forces F and F,. The re- 
sultant P" of P and F will 
meet the resultant P", of P x 
and P', in some point D, 
since (Art. 16) P'^ makes a 
less angle with P a than P" 
does with P. We shall then 
have, as before, the four forces 
P, Pj, P', and P',, acting at 
D, of which P' and F, are equal and opposite, while P and P 2 
are opposite and unequal, and have a resultant R=Pj— P, act- 
ing at any point in its direction, as C in AB produced. 

To determine the point C, we have (Art. 22), from the tri- 
angles ACD and BCD, as in Case 1°, 

P : F =DC : CA 
and F^P^BCrCD; 

compounding P : P^BC : CA, or P.AC-P^CB. 

37. Def. When AB is perpendicular to the direction of the 
forces, AC and BC are called the arms of the forces, and the 
products P.AC and P a .CB are the moments of the forces about 
the point C. 

38. Cor. If to the point C we apply a force R,, equal and 
opposite to R, the forces P, P 1? and R l will be in equilibrium 
about that point. For the resultant R, passing through C, will 
be counteracted by R , , so there can be no motion of transla- 
tion. And if we draw through C a line perpendicular to the 




PARALLEL FORCES. 17 

direction of the forces, calling the parts intercepted by P and 
Pj respectively p and p x ; from the similar triangles, thus 
formed, we should have 

BCiCA^p, :p. 

Hence the moments of P and P,, which measure the tendency 
to rotation in opposite directions, being equal, there can be no 
motion of rotation. 

39. Prop. To determine the point of application of the re- 
sultant in terms of the components and distance between their 
points of application. 

1°. When the forces act in the same direction. By Art. 36 
and figure, we have 

P.AC=P 1 .BC=P 1 .(AB-AC), 

or AC= F ?^-.AB; 

and P.AC=P 1 .BC=P(AB-BC), 

or BC= F ^-.AB. 

2°. When the forces act in opposite directions, we have 
P.AC=P I .BC=P 1 .(AC-AB), 

or AC^p^Lp.AB; 

P 

and similarly, BC = p p.AB. 

40. Cor. 1. When the forces act in opposite directions, the 
resultant lies without the components and on the side of the 
larger. 

41. Cor. 2. When the forces are equal and opposite, we 
have 

R=P 1 -P=0, 

or there will be no motion of translation. 

Also, AC=p^AB=^-.AB=ao. 

B 



18 



STATICS. 



Hence an equilibrium can not be produced except by the ap- 
plication of an infinitely small force at a point whose distance 
is infinite ; that is, two equal and parallel forces, acting in op- 
posite directions, can have no single resultant. 

42. Def. Such forces are called a statical couple. Their ef- 
fect is, tendency to rotatory motion only, and all tendency to 
rotatory motion can be referred to forces forming such couples. 

43. Prop. To find the resultant of any number of parallel 
forces acting at any points in the same plane. 




1°. Let the parallel forces P,, P 2 , P 3 . . . P w all act in the 
same direction. Draw any two rectangular axes OX, OY, 
and let 

re, y j be the co-ordinates of A, the point of application of V x 
x 2 y 2 " " B, - « " P 2 

x 3 y 3 « - D, " " " P 3 



x n y n 



x' y' 

x" y" 



N, 

C, 

E, 



" P* 

" R, 
" R. 



x y " " R, " " " R- 

Draw Ac and Cb parallel to OX. The triangles ACc and 
CBb are similar, and give 

AC : BC=Cc : Bb. 
By Art. 36, R^P.+P^ 



PARALLEL FORCES. 



19 



and P, : P 2 =BC : AC=Bb : Cc, or ~P,.Cc=¥ 2 .Bb, 
P 1 .(CH-AF)=P 2 .(BG-CH), 
(Bi+P^CH^P^F+P.-BG; 

or Ri-y=Pi-yi+P»-3r a - 

Compounding R 2 with P 3 , we shall find the second result- 
ant, 

R a =R 1 +P 3 =P 1 +P a +P 3 , 
and R a .EL=R 1 .CH+P 3 .DK, 

R 2 .EL=P T AF+P 2 .BG+P 3 DK; 

or »^-y"=P.y.+P.S!it^,- 

By continuing the same process, we should find, ultimately, 
R=P 1 +P a +P 3 +...P„ (a), 

Ry=P 1 y 1 +P,y a +P 3 y,+ ....P-y« (4 

By drawing lines from A, C, B, E, D, &c, parallel to OX, 
we may find, similarly, 

Ri=P 1 z 1 +P 2 x 2 +P 3 a; 3 +, &c +Y n x n (b). 

2°. When some of the forces act in opposite directions. To 
these the negative sign must be prefixed. 

If Rj, the resultant of y 
Pj and P 2 , as found in the 
previous case, be com- 
pounded with P 3 acting 
at D, in a direction oppo- 




site to that of Pj and P 2 , 
by joining CD and pro- 
ducing it (Art. 40), in the 
direction of the greater 
force, say R,, we have 
R 2 , the second resultant. 

R 2 =R a — P 3 , and E being its point of application, Rj.EC 
=P 3 .ED. Drawing Ce, Ed, parallel to OX, and meeting DK 
in d and LE produced in e, the triangles ECe, EDc? are simi- 
lar ; and 

EC : DE=Ee : T)d. 
.-. R^E^P^Drf; 
or R 1 (CH-EL)=P 3 .(DK-EL), 



20 



STATICS. 



or (R 1 -P 3 ).EL=R 1 .CH-P 3 .DK, 

or R a .EL=P 1 .AF+P a BG"P 3 .DK, 

or R.-y"=P,y,+P,y a -Psys> 

and so on for any other forces acting in directions opposite to 
P, and P 2 . 

3°. When the point of application of either force lies in any 
other angle than YOX, to its co-ordinates must be given their 
appropriate sign. 

The ordinate y 2 of B, the point of application of P 2 , will be 
negative. Draw Aa, Cb parallel to OX. 




H \G 

P 1 .CA=P a .CB. 

By sim. triangles, P 1 .Ca=P 2 .B&; 

or P 1 (AF-CH)=P a (CH+GB) > 

or (P 1 +P 2 ).CH=P 1 .AF-P 2 .BG, 



or 



R.^P.y.-P.y: 



.*. The general formulae (a), (b), (c) will apply to all cases, 
by giving the proper signs to the forces and to the co-ordinates 
of their points of application. 

These formulae are more concisely written by using the 
Greek letter 2 as the sign of summation, P to represent either 
force, and x, y its co-ordinates ; thus, 

R-=S(P) (a'), 

R..i=2(P.z) (&'), 

R.y=s(P.y) (C). 

When, therefore, the magnitudes of the forces and the co-or- 



PARALLEL FORCES. 21 

dinates of their points of application are given, (a) will give 
the magnitude of the resultant, and (b) and (c) the co-ordinates 
x, y of its point of application. 

44. Def. The point whose co-ordinates are x, y is called the 
center of parallel forces. Its position depends on the magni- 
tude of the forces and the co-ordinates of their points of appli- 
cation, but is independent of their common direction; for, by 
turning the forces around their respective points of applica- 
tion, at the same time preserving their parallelism, it will be 
seen that the position of the point E is not thereby affected. 

45. Prop. The moment of the resultant of any number of par- 
allel forces, acting in the same plane, is equal to the algebraic 
sum of the moments of the components. 

Since the origin of the co-ordinate axes and their direction 
are arbitrary (see figure of Art. 43), suppose the axis OX to be 
drawn perpendicular to the direction of the forces, and the 
forces produced to intersect it. Then V 1 x 1 , V 2 x 2 , &c, will 
represent the moments of P a , P 2 respectively, and Rx the mo- 
ment of the resultant. 

46. Def. The moment of a force with reference to a plane is 
the product of the intensity of the force by the perpendicular 
let fall from the point of application upon the plane. Thus 
Pj^j is the moment of the force P,, in reference to the plane 
passing through OY perpendicular to OX, and P^ is the 
moment of the force P,, with reference to the plane through 
OX perpendicular to OY. 

47. Prop. To determine the conditions of equilibrium of any 
number of parallel forces. 

Suppose the forces in the figure (Art. 43) all turned round 
their points of application so as to become perpendicular to the 
plane of YOX. They will no longer be in the same plane, but 
as their intensity is not thus changed, nor their points of appli- 
cation, (a'), (b'), (c 1 ) will still hold. Then, 
1°. We must obviously have R = in (a'), or 

P,+P 2 -fP 3 +, &c. = 0. 
2°. If this value of R be substituted in (&'), we have 
__2.Fx_2.Fx 
X ~ R " 0~~' 



22 STATICS. 

Therefore x will be infinite unless 2.P:c=0. But when x is in- 
finite, we have a couple {Art. 41), and, consequently, there can 
be no equilibrium. Hence we must also have S.Px=0, or 

P 1 z 1 +P 2 £ 2 +P 3 Z3+, &c.=0. 
And in the same manner it will appear that S.Py— in case of 
an equilibrium. But 2.P# and S.Py are the moments of the 
forces in reference to two planes parallel to their directions. 
Hence the necessary conditions of equilibrium are, 

1°. The sum of the forces must be equal to zero. 
2°. The sum of their moments, with reference to two planes 
parallel to their directions, must each be equal to zero. 

48. Cor. 1. If R=0, but 2.P.Z and 2.P?/ are not equal to 
zero, there will be no motion of translation, but simply a mo- 
tion of rotation. 

For if Rj be the resultant of all the positive forces, and x t 
the abscissa of its point of application, R 2 the resultant of all 
the negative forces, and x 2 its abscissa, 

then RjXj- R 2 ^ 2 ^0; 

or, since R,=R 2 , R l (x l —x 2 )'^.0. 

But R l being finite, x r —x 2 must also be finite. 

That is, the points of application of R l and R 2 are not the 
same, and a tendency to motion of rotation around the axis of 
Y exists. 

And S.Py^O 

gives, in like manner, a tendency to motion around the axis 
of x. 

49. Cor. 2. If the equilibrium subsist, one force, as P,, must 
be equal and opposite to the resultant of all the others. 

50. Cor. 3. If the equilibrium subsist with one direction, it 
will subsist whatever be the common direction of the forces. 

51. EXAMPLES. 

1. Two parallel forces, acting in the same direction, have 
their magnitudes 5 and 13, and their points of application A 



PARALLEL FORCES. 23 

and B 6 feet distant. Find the magnitude of their resultant, 

and its point of application C. 

Ans. R=18, 
AC=4J, 
BC=lf. 

2. Find the resultant, and its point of application, when the 
forces in the last question act in opposite directions. 

Ans. R=8, 
AC=9f, 
BC=3|. 

3. If two parallel forces, P and Q, act in the same direction 
at A and B, and make an angle 6 with AB, find the moment 
of each about the point of application of their resultant. 

P.Q 

Ans. p , q AB sin. 0. 

4. If the weights 1, 2, 3, 4, and 5 lbs. act perpendicularly 
to a straight line at the respective distances of 1, 2, 3, 4, and 5 
feet from one extremity, required their resultant and its point 
of application. 

Solution. From equations (a) and (b) we have 

R=2.P=l + 2+3+4+5=15 lbs. 

And, taking the extremity of the line for the origin of co-or- 
dinates, 

Rx=15.x=I,.Vx=lX 1+2X2+3X3+4X4+5X5=55. 
,\ a; =3 J feet. 

5. Let the weights 4, — 7, 8, and —3 lbs. act perpendicu- 
larly to a straight line at the points A, B, C, and D, so that 
AB=5 feet, BC=4 feet, and CD=2 feet; find the resultant 
and its point of application E. 

Ans. R=2 lbs., 
AE=2 feet. 

6. Let three forces which, if concurring, would be in equi- 
librium, act each in the side of the triangle which represents 
them in magnitude and direction. Show that they are equiv- 
alent to a statical couple. 



CHAPTER III. 



THEORY OF COUPLES. 

52. Def. A statical couple consists of two equal and parallel 
forces acting in opposite directions at different points of a body. 

53. Def. The arm of a couple is the perpendicular distance 
between the directions of the forces. , 

54. Def. The moment of a couple is the product of the arm 
by one of the forces. 

55. Prop. A couple may be turned round in any manner in 
its own plane without altering its statical effect. 

Let PjABP 2 be the original couple. Suppose the arm AB 
p 2 turned around any 

f -p s point in it to the posi- 

tion ab. Apply the 
equal and opposite 
forces P 3 and P 4 per- 
pendicularly to ab at 
a, and similarly P 5 
and P 6 at b, and let 
each be equal to P x 
or P 2 . These forces, 
being in equilibrium, 
will not disturb the 
system. The two equal 
forces Pj at A, and P 4 at a, may be regarded as acting at their 
point of intersection E, and will have a resultant in the direc- 
tion CE bisecting the angle P,EP 4 . Similarly, the resultant 
of P 2 and P 6 at D will be an equal force in the direction CD. 
These forces, being equal and opposite, may be removed; 
that is, we may remove from the system the forces P^ P 2 , P 4 , 
and P 6 , and we have remaining the forces P 3 and P 5 at a 




COUPLES. 



25 




and b, forming the couple P 3 afrP 5 , which is the same as if we 
had turned the original couple round the point C until its arm 
came to the position ab. 

56. Prop. A couple may be removed to any position in its own 
plane, the parallelism of its arm being preserved, without chang- 
ing its statical effect. 

Let the arm AB of the couple P^BP,, be removed in its 
own plane to the parallel po- 
sition ab, and let the forces 
P 3 , P 4 , P.. P 6 , each equal to 
the original forces, be applied 
perpendicularly to ab, at the 
extremities a and b, in oppo- 
site pairs. 

Join Ab and aB. These 
lines will bisect each other 
in C. Pj at A and P 5 at b 
will have a resultant 2P X at 
C, parallel to the original direction. Similarly, P 2 at B and 
P 4 at a have a resultant =2P 2 at C, opposite to the former; 
these will consequently balance each other, and may therefore 
be removed, or the forces P a , P 2 , P 4 , P 5 may be removed, 
and we have remaining the couple P 3 a&P 6 , equivalent to the 
original couple, removed parallel to itself in its own plane. 

57. Prop. A couple may be removed into any other plane, par- 
allel to the original one, without altering its statical effect, the 
parallelism of its arm being preserved. 

Let the arm AB of the original couple be transferred from 
its own plane MN parallel to itself to ab in the parallel plane 
RS. Let forces be applied at a and b, as in the preceding 
proposition, each being equal to P, or P 2 . Join Kb and aB, 
These lines will bisect each other in C. The forces P T and 
P 6 will have a resultant = 2P X at C, and P 2 and P 4 a result- 
ant = 2P 2 at C. These resultants will be equal and opposite, 
and may therefore be removed without disturbing the statical 
effect of the system. We have then remaining the couple 



26 



T A T I C S. 




V 3 abV 5 , equivalent to the original couple, and transferred to 
the plane RS. 

58. Prop. All statical couples are equivalent to each other 
whose planes are parallel and moments equal. 



Let 



Q4 



P.ABP, 



0? 



and Q^EQ, 



E 



AP2 



Qs 



Pi=P+Qi 



be any two couples whose 
planes are parallel and 
moments equal. If they 
are not in the same plane, 
by the foregoing propo- 
sitions let the latter be 
transferred to the plane 
of the former, and moved 
in that plane until the 
extremities D and A of 
their arms coincide, and 
DE take the position 



i 



AC. Apply at C, perpendicularly to AC, two equal and op- 
posite forces Q 3 and Q 4 , each equal to Q 1 . Resolve V l into 
two forces F and Q 1? so that P 1 =P / +Q 1 , or P 1 -Q 1 =F. 

Then, by hypothesis, 



COUPLES. 27 

P 1 .AB=Q 1 .DE=Q 1 .AC=Q 1 .AB.+Q 1 .BC=Q 1 .AB+Q 3 .BC 

and (P 1 -Q 1 )AB=F.AB=Q S .BC, 

or the resultant of P' and Q 3 (Art. 36) passes through B. It 
is also equal and opposite to P 2 . Hence the forces P', Q 3 , 
and P 2 may be removed, and there will remain the couple 
QjACQ 4 , equivalent to P^BP^ 

59. Prop. Any statical couple may be changed into another, 
which shall be equivalent and have an arm of given length. . 

Let P 1 jo 1 be the moment of "any couple, of which Pj is one 
of the forces and p , its arm, and let p be the given arm. Find 
a fourth proportional P^ to p, p l9 and P, ; or take p :p 1 = 
P, : P'j, which gives ?' 1 p=Y 1 p l . Hence, by Art. 58, their 
moments being equal, and in the same plane, the couples are 
equivalent. 

60. Def. The axis of a couple is a line perpendicular to the 
plane of the couple. 

If the length of the axis be taken proportional to the mo- 
ment of the couple, and drawn above its plane when the couple 
is positive or tends to produce rotatory motion in the direction 
of the hands of a watch, and below its plane, when negative 
or tends to produce motion in the opposite direction, then the 
axis will completely represent the couple in position, intensity, 
and sign. 

61. Cor. By considering the previous propositions, it will be 
obvious that the axis of a couple, as thus defined and limited, 
may be removed parallel to itself, to any position within the 
body acted on by the couple. 

62. Def. The resultant of two or more couples is one which 
will produce the same effect singly as the couples themselves 
jointly. 

63. Prop. The moment of the resultant of two or more couples 
in the same or parallel planes, equals the algebraic sum of the 
moments of the component couples. 

Let P 1? P 2 , P 3 , &c, be the forces ; p 19 p 2 , p s , &c, their 
arms respectively. The couples may all be removed into one 



28 



STATICS. 



A p * 



V"Pi 
VP* 

VPi 



plane (Art. 57), turned round in that 
plane (Art. 55), moved in it (Art. 56), 
and their arms changed (Art. 59) to a 
common arm, while their moments re- 
main the same. Let p be the com- 
mon arm AB ; P',, P' a , P' 3 , &c, the 
forces; so that V\p = Y l p i , Y' 2 p = 
V 2 p 2 , P' 3 p=P 3 p 3 , &c. Now the 



forces F 1 ,P / 2 ,F ; 
lent to a force 



at A 



are equiva- 



P'j+P'a+P'g+.&C, 



P P 



+ 



P 3 /> 



We have 




And the forces at B are equal to the same sum. 
then for the moment of the resultant couple, 

(P / 1 +P , a +P 3 + > &c.) AB=(F 1 +F a +F 3 +)jo, 

-P 1 p 1 +P^ 2 +P 3 p 3 +'&c. 

If either of the original 
couples, as P 3 jo 3 ,tend to pro- 
duce motion in the opposite 
direction, its sign will be neg- 
ative. The sum of the forces 
will be 

F 1+ F 2 -F 3 +,&c, 
and the moment of the result- 
ant couple 
(P' 1 +P' 2 -P' 3 +,&c.)jo=P^ I +P 2 p 2 -P D? 3+.&c. 

64. Cor. The axis of the resultant couple will be equal to 
the algebraic sum of the axes of the component couples. 

Schol. The composition of couples in the same or parallel 
planes, by means of their axes, is therefore analogous to the 
composition of conspiring forces. 

65. Prop. To find the resultant of two couples in different 
planes inclined to each other. 

Let each couple be turned round and moved in its own 
plane until its arm coincide with the common intersection of 
their planes, and let the arms be made coincident and of the 



COUPLES. 



29 




same length, the moments 
of the couples remaining 
the same. Let AB be this 
common arm, and PABP, 
QABQ the couples. Com- 
plete the parallelograms 
on the lines representing 
P and Q, and the diago- 
nals will represent their 
resultants R, which may 
be substituted for them. 
Hence we have the result- 
ant couple RABR equiva- 
lent to the component couples PABP and QABQ. 

66. Cor. The diagonal of the parallelogram, constructed on 
the lines representing the axes of the component couples, will 
represent the axis of a couple equivalent to them. 

Let OK and OL be the 
axes of the couples, and 6 the 
angle made by the planes of 
the couples. Since the axes 
are perpendicular to these 
planes, the angle made by the 
axes will be 6. We have, 
therefore, 0=PAQ=LOK. 

By the triangle of forces, 

R 2 =P 2 +Q 2 +2PQ cos. e. 
.-. R.AB 




ABv , P 2 +Q 2 +2PQcos. 0, 



.(since OK : 



= VP 9 .AB 9 + Q a .AB 9 +2P.ABxQ.ABcos.0; 
P.AB, &c), 

= v , OK 2 +OL 2 +20K.OL cos. d, 
=OM. 

And OK, OL being respectively perpendicular to the planes 
of the component couples, OM will be perpendicular to the 
resultant couple. 

67. Schol. The above is analogous to the parallelogram of 
forces, and may be called the parallelogram of couples. 



30 STATICS. 

If L and M represent the axes or moments of the compo- 
nent couples, and G the axis or moment of their resultant, then 
G 2 =L 2 +M 2 +2LM cos. 0. 

If L, M, N were the axes of three component couples, it 
might be shown that G, the axis of the resultant couple, would 
be represented in magnitude by the diagonal of the parallelo- 
piped formed upon them. 

If the planes of the three couples are at right angles each to 
the other two, or if L, M, and N are at right angles each to 
the planes passing through the other two, the parallek>piped 
would be rectangular, and we should have, 
G 2 =L a +M 2 +N 2 . 



CHAPTER IV. 

ANALYTICAL STATICS IN TWO DIMENSIONS. 

In this chapter the points of application and the directions 
of the forces will be referred to co-ordinate axes at right an- 
gles to each other. 

68. Prop. To find the magnitude and direction of the result- 
ant of any number of concurring forces in the same plane. 

Let P l5 P 2 , P 3 , &c. . . P„ be the n forces, and let the point 





A 


^ Pl 








o 


- a 


I X 



at which they act be taken for the origin of co-ordinates. 
Let P, make the angle <z a with OX, and j3, with OY. 

Po " «o " 0o 



&c, 



&c, 



£3 

&c, 



If OA represent the force P 1? and the parallelogram OMAN 
be completed, OM will be the component of P, in OX, and 
ON that in OY. And 



32 STATICS. 

OM=X 1 =P 1 cos. a l9 ON=Y 1 =P 1 cos. f3 u 

Pursuing the same course with the forces P 2 , P 3 , &c, and 
calling the resolved parts of the forces respectively in OX, 

X 2 , X 3 , X 4 , &c. . . . X M , 
and those in OY Y 2 , Y 3 , Y 4 , &c. . . . Y M , 
we shall obtain 

X,=Pj cos. a 19 Y 1 =P 1 cos. (3 l9 

X 2 =P 2 cos. a 2 , Y 2 =P 2 cos - / 3 2» 

X 3 =P 3 cos. a 3 , Y 3 =P 3 cos. 3 , 

&c, &c, &c, &c, 

X n =P„ cos. a n , Y n =P„ cos. (3 n . 

But the components in OX are equivalent to a single force, 

=X 1 +X 2 +X 3 +, &c X B =2.X, 

and those in OY to a single force, 

=Y 1 +Y a +Y 3 +, &c. . . . Y W =S.Y, 
the Greek letter 2 being used as the sign of summation. 
Hence we have 

S.X=2.P cos. a, (1) 

2.Y=2.P cos. )3. (2) 

Now if R be the resultant required, and the angle it makes 
with OX, the resolved parts of R in the axes must equal the 
resolved parts of the forces in the same directions. 

.-. R cos. 0=2.X, (3) 

R sin. 0=2.Y. (4) 

Hence Tan. 0=^. (5) 

Also, R 2 . cos. 2 ^+R 2 .sin. 2 0=R 2 =(S.X) 2 + (S.Y) 2 ; 



or R=v , (2.X) 2 + (^.Y) 2 . (6) 

Equations (1) and (2) give the values of 2.X and S.Y, by 
which we obtain from (5) and R from (6) ; or R may be ob- 
tained, numerically, more readily from the equation 
R cos. 0=2.X. 

T. 2.X 

For R= 2T=2.Xsec.0. (?) 

COS. V 

69. Schol. It is readily seen that the signs of the compo- 



ANALYTICAL STATICS. 



33 




nents along either axis are involved in the trigonometrical ex- 
pressions of their values, (1) and 
(2). Assume the directions OX 
and OY positive, and OX', OY' 
negative. The components 0^ 15 
Oy x of P x are positive ; their 
values, P a cos. a l9 P x cos. j3,, 
will also be positive, since the 
cosines of angles in the first 
quadrant are positive. The com- 
ponents of P 2 , which lies in the 
angle X'OY, are Ox 2 , Oy 2 , the 
former negative and the latter 
positive. The value of Ox 2 = 
P 2 cos. a 2 is negative, since a 2 , reckoned from OX around to 
the left, will be in the second quadrant ; and that of 0y 2 =V 2 
cos. (3 2 positive, since (3 2 , reckoned from OY around to the 
right, will be in the fourth quadrant ; or, if reckoned from OY 
around to the left, will be negative, and less than 90°. 

The components Ox 3 , Oy 3 of P 3 , in the angle X'OY', are 
both negative, and their values P 3 cos. a 3 , P 3 cos. (3 3 are neg- 
ative, because the angles a 3 and (3 3 are both in the third quad- 
rant. The components Ox 4 , Oy 4 of P 4 , which lies in the 
angle Y'OX, are, the former positive and the latter negative. 
The value of 0^ 4 =P 4 cos. a 4 is positive, because <z 4 is in the 
fourth quadrant, or, if taken negatively, is less than 90° ; that 
of Oz/ 4 = P 4 cos. j3 4 is negative, because /3 4 is in the second 
quadrant. By the above mode of reckoning the angles, we al- 
ways find a-\-(3=90°, or one the complement of the other. 

70. Prop. Required the conditions of equilibrium of any 
number of concurring forces in the same plane. 

Since the forces are in equilibrium, their resultant must be 
equal to zero, or R=0. This gives, by (6), 

(2.X) 2 +(2.Y) 2 =0. 
But each term being a square, is essentially positive. The 
equation, therefore, can only be satisfied by making each term 
equal to zero at the same time. 

C 



34 



STATICS. 



Hence 2.X=0, 2.Y=0, (8) 

are the two necessary and sufficient equations of equilibrium of 
any number of concurring forces in the same plane ; that is, 
the sums of the components of all the forces resolved in any 
two rectangular directions, must be separately equal to zero. 

71. Prop. Required the expressions for the resultant force 
and resultant couple of any number of forces acting at different 
points in the same plane. 

Let Pj, P 2 , &c. . . . P n be any number of forces in the same 
plane. Take the co-ordinate axes OX, O Y in the plane of the 
forces, and let a,, a 2 , a 3 . . . a n be the angles the forces make 
respectively with OX, and P l9 j3 3 , j3 3 . . . fi n the angles they 
make with OY. 

Let x 19 y x be the 
co-ordinates OM a , 
M^! of A j, the point 
of application of the 
force P,, and x 2 . y 3 , 
&c. . . . x n , y n be the 
co-ordinates of the 
points of application 
X of the others respect- 
ively. Resolving P, 
in the directions of 
OX and OY, we shall 
have for the components 

X 1 =P 1 cos. a 19 Y 1 =P 1 cos. j3j, 

at the point A , . 

Now apply at O, in OX, two equal and opposite forces, each 
equal to X, ; this will not affect the system. Then, instead 
of the single force X x at A 15 we have X a at A 1? or N, and 
— Xj at O, which together form a couple, and X x at O. That 
is, for the force Xj at A,, we may substitute 

X T at O, and the couple X 1 .ON 1 =X 1 y 1 . 
In like manner, applying at O, in OY, two forces, opposite and 
each equal to Y v , we should have Y 2 at A x equivalent to 
Y, at O, and the couple -Y.OM^-Y.^. 




ANALYTICAL STATICS. 



35 



This last couple will be negative, since it tends to produce ro- 
tation in the contrary direction to that of the other. 

By pursuing the same course with each of the other forces. 
we should obviously have acting at O, in OX, a sum of forces 

X 1 +X 2 +X 3 +,&c X n =S.X, (9) 

and at O, in OY, 

Y 1 +Y 2 +Y 3 +,&c Y n =2.Y, (10) 

and the couples 

X 1 y 1 +X 2 y 2 +X 3 ?/3 .... X„3/ n =2.Xy, (11) 

-Y 1 ^ 1 -Y 2 ^ 2 -Y 3 ^3 -Y n x,=2.-Yx. (12) 

We have now reduced the whole to a system of concurring 
forces and a system of couples. The couples being in the 
same plane, the moment of- the resultant (Art. 63) will be 
equal to the algebraic sum of the moments of the components, 
or the resultant axis G'will be equal to the algebraic sum of 
the component axes. Hence, taking the sum of (11) and (12), 
G=2.Xy-Z.Yx=2.{Xy-Yx). (13) 

Now if R be the resultant force acting at O, and the angle 
it makes with OX, we have, as before, 
R cos. 6=i.X, 
R sin. d=2.Y. 

2.Y 
Tan. 0=^, (14) 

and R 2 =(2.X) 2 +(2.Y) 2 . (15) 

72. Schol. 1. The equations (13), (14), and (15) determine 
the magnitude of G and the magnitude and direction of R. 
To construct these insults, draw 
through the origin O the line OR, 
making the angle 6 with OX, to 
represent R. Then put G= 
l.(Xy—Yx)='Rr, by which the 
moment of the resultant couple 
is changed into another, whose 
forces are each equal to R, and 
arm equal to r. Let this be 
moved and turned round until 
one of its forces acts at O in an 




36 STATICS. 

opposite direction to the resultant force ; draw OA perpendic- 
ular to OR and equal to r, and AR' parallel to OR. Then 
R'AOR' represents the resultant couple. The two forces at O, 
being equal and opposite, may be removed, and we have the 
final resultant acting in AR', which makes the angle 6 with 
OX. 

73. Schol. 2. To find the equation of this final resultant. It 
will be of the form 

y=ax+b. 

s.Y 

But «= tan. 6= 



and b=OB 



2.X' 

OA r G G 



cos. 6 cos. R cos. 6 2.X* 

Therefore, by substitution, we have for the equation 

v v G 

" = ET* + IS (16)> 

74. Prop. To determine the conditions of equilibrium of any 
number of forces acting at different points in the same plane. 

In order that there may be no motion of translation, we 
must have R=0, which gives, as before, 

2.X=0, (17) 

and 2.Y=0. (18) 

And, in order that there may be no motion of rotation, we 
must also have 

G=2(Xy-Ya:) = 0. (19) 

Equations (17), (18), (19) are the three necessary and suffi- 
cient conditions of equilibrium. 

75. Cor. When there is a fixed point in the system, if this 
point be taken for the origin, its resistance will destroy the ef- 
fect of the resultant force R, and the sole condition of equilib- 
rium will then be G=0, 

or l.(Xy-Yx)=0; 

or there must be no tendency to rotation around the fixed 

point. 



ANALYTICAL STATICS. 



37 



ECIUILIBRIUM OF A POINT ON A PLANE CURVE. 

76. If a point be kept at rest on a plane curve by the action 
of any number offerees in the plane of the curve, the resultant 
must obviously be in the direction of the normal to the curve 
at that point, and equivalent to the pressure the curve sustains. 
For, if the resultant had any other direction, it might be re- 
solved into two, one in the direction of the normal, and the 
other in the direction of the tangent to the curve ; the former 
would be opposed by the reaction of the curve ; the latter, be- 
ing unopposed, would cause the point to move. Hence, 

77. Prop. To determine the conditions of equilibrium of a 
point, retained on a plane curve or line, by forces acting in its 
plane. 

Let N be the normal force of reaction of the curve, and a 
the angle made by the normal with the axis of x. Also, let 
2.X, 2. Y be the sums of the components of all the other forces 
resolved parallel to each axis respectively, and R their result- 
ant. The resistance N may be considered a new force, which, 
together with the other forces, retains the point in equilibrium 
independently of the curve. If, therefore, N be resolved in 
the direction of each axis, we have (8) 

N cos. a+2.X=0, (a) 

N sin. a+2.Y = 0; (b) 

or, transposing, squaring, adding, and reducing, 

N= V(2,X) 2 +(S.Y) 2 =R, 
or the reaction is equal to the resultant of all the other forces. 

Now let i be the inclination r y 

to the axis of x, of the tangent 
to the curve through the point, 
then a=90°-\-i. 




and 



cos. a=cos. (90°+i) = — sin. i, 
sin. «=cos. i. 



Substituting these values of cos. a and sin. a in (a) and (b), and 
dividing, we get 



38 STATICS. 

2 X 
Tan. i=--^, or 2.X+2.Y tan. i=0. (20) 
_'. j. 

But the differential expression for the tangent of the angle 

dxi 
which the tangent makes with the axis of x is -A Therefore, 

substituting and reducing, 

2.X.dx+2.Y.dy=0. (21) 

Whenever the line on which the point is retained is right, 

(20) may be used ; but if the line be a curve, (21) will, in gen- 

dy 
eral, be necessary, and -j- must be deduced from the equation 

of the given curve. 



VIRTUAL VELOCITIES. 

78. Def. If any forces P a and P 2 act at the point A, and 
this point be displaced through an In- 

1 definitely small space Aa, and the 
perpendiculars ap 1 , ap 2 be drawn 
from a on the directions of the forces, 
#2 then Ajo, and Ap 2 are called the vir- 

tual velocities of the forces V 1 and P 2 ; Ap l9 measured in the 
direction of the force P,, is positive, and Ap 2 , measured in the 
direction of P 2 produced, is negative. 

79. The principle of virtual velocities is thus enunciated : 

If any number of forces be in equilibrium at one or more 
points of a rigid body, then if this body receive an indefinitely 
small displacement, the algebraic sum of the products of each 
force into its virtual velocity is equal to zero. 

80. Prop. To prove the principle of virtual velocities for 
concurring forces in one plane. 

Let A be the point at which the forces P 1? P 2 , P 3 , &c 

P n act ; a x , a 2 , a 3 , &c. . . . a n the angles they make respect- 
ively with OX, and j3 1? j3 2 , (3 5 . . . ft, the angles they make 
with OY. Let 6 be the angle which the direction of the dis- 
placement Aa makes with OX. 




ANALYTICAL STATICS. 



39 




Let v 19 v 2 , v 3 , &c. . . , v n be the virtual velocities of the 
forces respectively. 
Then 
v l =Ap l =Aa cos.p 1 Aa=Aa. cos. («,— 0), 

=Aa(cos.a 1 cos. 0+sin. a, sin.0), 
and ¥ l v 1 =~P l .Aa(cos.a 1 cos.^+sin.flj sin.0), 

= Aa(cos. 0.P , cos. a , +sin. 0.P , sin. a x ). 
In like manner, for P 2 we should have 

P 2 v 2 = Aa(cos. 0.P 2 cos. a 2 +sin. 0.P 2 sin. « 2 ), 
and so for the other forces ; and taking their sum, we should 
get, remembering that sin. a=cos. /3, 

S.Pw=P 1 » 1 +P a v a +P 3 W3+ > &c P n v„, 

=Aa[cos. (Pj cos. a, +P 2 cos. a 2 +, &c P„ cos. a n ) 

+ sin. 0(P 1 cos.^+P,, cos.0 2 +,&c P n cos.0 n )]. 

But when there is an equilibrium at a point (8), 

Pj cos.a 1 +P 2 cos.a 2 -fP 3 cos. <2 3 +,&c P„cos.a„=2.X=0, 

and 

P, cos. j3, +P 2 cos.0 2 +P 3 cos.0 3 +,&c. . . . P n cos.j3 B =S.Y=0. 

Hence 2.Pv=0, 

or the principle is true when the forces all act at a point. 

81. Prop. To prove the principle of virtual velocities for 
forces acting at different points in the same plane. 

The points are supposed to be invariably connected by rigid 
lines or rods without weight, which transmit the actions and 
reactions of the particles or points upon each other. 



40 STATICS. 

Let A,, A 2 , A 3 , &c. . . . A n be the particles to which the 

forces Pj, P 2 , P 3 , &c. . . . P n are applied, v l9 v 2 , v 3 . . . v n the 

virtual velocities of the forces respectively. 

Let r aj a a be the action of the particle A 1 upon the particle A 2 , 

ra 2 a x " reaction of " A 2 " " A l5 

r ai a, " action of " A, " " A 3 , 

r a a, " reaction of " A , " " A,. 

31 •> * 

&c, &c, &c, 

Let Va x a 2 , Va^, va x a 3 , Va 3 a l9 &c, &c, be the corresponding 

virtual velocities. 

Then Va x a=Va 2 a x , Va x a 3 =Va 3 a l9 &c, &c, from the nature of 

action and reaction. 

Also, ra x a= — ra 2 a x , r ai a 3 = — ra 3 a l9 &c, &c. For, let Aj 

R ^ AiP^ A2P2 ^Jt 2 

and A 2 be the particles displaced to a l and a 2 . Draw the 
perpendiculars a 1 p 1 , a 2 p 2 . Then, if the line a l a 2 is par- 
allel to AjA^j, it is obvious that A 1 p 1 =A 2 p 2 . But A^ is 
the virtual velocity of R l5 and A 2 p 2 of R 2 , and they have op- 
posite signs. 



If a l a 2 is not parallel to AjA 2 , let them meet when pro- 
c A-1P1 A2P 2 



duced in some point C. Since the displacements are indefin- 
itely small, the perpendiculars a 1 p i ,a 2 p 2 coincide with cir- 
cular arcs whose center is C, and Ca J =Cp l , Ca 2 =Cp 2 . 
But A.p^Cp.-CA^Ca.-CA,, 
and A 2 p 2 =Cp 2 —CA 2 =Ca 2 ~CA 2 = (Ca 1 +a 1 a 2 ) — (CA l + 

= Ca 1 -CA 1 .. 
.'• A,^, = A 2 p 2 , 

Or Va x a=— Va 2 a x . 

Let the sum of the products of all the forces P 15 P 2 , &c, 
into their virtual velocities, acting on the particle 



ANALYTICAL STATICS. 41 

A a be 2.(P ai .v a ) 9 

those on A 2 be S.(Pa a .ua a ), 

those on A 3 be 2.(P« 3 .i^ 3 ), 

&c, &c, 

those on A„ be ^.(?a n .v a ). 

Since each particle is in equilibrium from the action of the 
external forces and the reactions of the others upon it, we 
have, by the last proposition, 

0='2.(Ya 1 .Va)- S rra i a 2 .Va l a 2 + ra 1 a 3 .Va 1 a 3 + , &C, 
= 2.(¥a 2 .Va 2 )+ra 2 a 1 .Va 2 a 1 + ra 2 a 3 .Va 2 a 3 J r, &C, 
= 2. (¥a 3 .Va 3 ) + ra 3 a 1 .Va 3 a 1 + ra 3 a 2 .Va 3 a 3 + , &C. , 

&C, &C, &C, 

= 2.(P« .Va)+r a a,.Va a+r a a.Va a+, &C. 

In taking the sum of the products for all the particles, the 
products of the reactions into their virtual velocities will dis- 
appear, being in pairs, equal in magnitude with contrary 
signs ; therefore we have 

S.(P« 1 .v« l )+5:.(P a8 .Ufl a )+i;.(Pa,.Wa3)+, &c 2.(P<v«O=0 ; 

or, generally, when there is an equilibrium, 

2.(Pd)=0. 

82. Prop. Conversely. If the sum of the products of the 
forces into their virtual velocities be equal to zero, or 2.(P.u.) 
=0, then there will be an equilibrium. 

For if the forces are not in equilibrium, they will be equiva- 
lent either to a single force or a single couple {Art. 74). 

In the first case, let R be the single resultant force ; then a 
force equal and opposite to R will reduce the system to equi- 
librium ; let u be its virtual velocity for any displacement. 
Since, with this new force, there will be an equilibrium, we 
have, by the preceding proposition, 

2.(Y.v)+R.u=0. 

But by hypothesis, 2.(P.u) = 0. .*. Rw=0, which, being true for 
all small displacements of the body, we must have R=0, or 



42 STATICS. 

the body was in equilibrium from the action of the original 
forces. 

In the second case, if the forces were equal to a resultant 
couple, it would be balanced by an equal and opposite couple. 
Let the forces of this opposite couple be Q, and Q/, and their 
virtual velocities for any displacement be q and q' respective- 
ly ; since they will reduce the system to equilibrium, we have, 
by the preceding proposition, 

2.(P.u)+Q?+Qy=0. 
But 2.(P.u)=0. .-. Q,q+Q,'q'=0, for all displacements, which 
is impossible, unless Q, and Q/ each equal zero, since they are 
parallel forces, and act at different points. 



CHAPTER V. 



THE CENTER OF GRAVITY. 



83. Def. Experiment shows that every particle of matter is 
subject to a force which attracts it in a direction perpendicu- 
lar to a horizontal plane or the surface of stagnant water. In 
reality, the directions of the forces, acting on several particles, 
meet nearly in the center of the earth ; but as this center is 

'very distant, compared with the distance of any particles con- 
sidered together, we may, without sensible error, regard their 
directions as parallel. 

This force is called gravity. 

84. Experiment shows, also, that the intensity of gravity 
varies in different parts of the earth's surface ; that it is least 
at the equator, and increases toward the poles in the ratio of 
the square of the sine of the latitude. It shows, also, that in 
the same latitude the intensity varies at different points in the 
same vertical line ; that it varies inversely as the square of the 
distance from the center. But for any points in the same sys- 
tem, or any bodies nearly in the same place, this variation of 
intensity, as well as difference of direction, may be neglected 
without error. Hence, 

85. Def. A heavy body is an assemblage of material points, 
or particles, acted on by equal parallel forces in the direction 
of the vertical to the earth's surface. 

86. Def. The weight of a body is the resultant of all the ef- 
forts which gravity exerts on its component particles. This 
resultant (Art. 43) is equal to their sum, and parallel to their 
common direction. 

87. Def. The mass or the quantity of matter of a body is 
the sum of all its component particles. 

88. Cor. If W represent the weight of a body, M the mass, 
and g the ratio of the intensity of gravity at any place, to its 



44 STATICS. 

intensity at another place where it is assumed as unity, we 
shall have 

W=g-M. (22) 

For the resultant of all the parallel actions of gravity on the 
particles of a body is equal to their sum, or the product of its 
intensity into the number. 

89. Def. The density of a body is the ratio of its mass to its 
volume ; or, if D represent the density, M the mass, and V the 
volume, 

D=y; (23) 

in which D, M, and V represent the number of units of each 
kind. 

90. Cor. Since W=#M, and M=VD, we have 

W=#VD. (24) 

91. Prop. The masses of two bodies of the same density are in 
the direct ratio of their volumes. 

If M, V, and D be the mass, volume, and density of one 
body, and ?n, v, and d the same of the other, by (23), 

M=VD and m=vd; v M : m=YD : vd ; (25) 
or, since D=d, M : m=Y : v. 

92. Prop. The masses being equal, the densities are inversely 
as the volumes. 

Since M : m=VD : vd; if M=m, YB=vd; 
or D : d=v : V. (26) 

93. Prop. The volumes being equal, the masses are directly as 
the densities. 

Making, in (25), Y=v, we have M : m=B : d. (27) 

94. Def. The center of gravity of a body is the center of the 
parallel forces of gravity on each of its component particles. 

95. Cor. Hence the determination of the center of gravity 
involves an immediate application of the doctrine of parallel 
forces, and we need only refer to results already obtained for 
many important deductions respecting the center of gravity. 



CENTEROFGRAVITY. 45 

1°. The resultant of all the vertical efforts of gravity on 
each of the elementary particles of a body passes through its 
center of gravity. (Arts. 44 and 94.) 

2°. This resultant is parallel to the forces ( Art. 36) ; that is, 
it is vertical, and its magnitude is equal to the weight of the 
body. 

3°. Whatever position we give to a body, this resultant will 
always pass through the center of gravity ; since changing the 
position is equivalent to changing the direction of the forces, 
without changing their points of application or parallelism. 

4°. A heavy body will be in equilibrium if its center of 
gravity be supported, whatever may be the situation of the 
body relative to the support, since, in this case, the resultant 
of the parallel forces of gravity will have a fixed point in its 
direction. 

5°. When we wish to find the center of gravity of several 
bodies, we can suppose the mass of each concentrated at its 
center of gravity, since the weight of each is a force propor- 
tional to its mass, and passing vertically through its center of 
gravity. Hence we have only to consider a system of heavy 
points. 

96. Def. A body is said to be symmetrical with respect to a 
plane when the lines joining its particles, two 
and two, are parallel, and bisected by the 
plane. 

Thus, let m, m' be two symmetrical par- 
ticles, so that the line mm' may be bisected 
in b by the plane aa'. Letting fall the perpen- 
diculars ma, m'a', the equality of the triangles 
mab, m'a'b gives ma=m'a'. Hence the par- 
ticles of a body, symmetrical with respect to 
a plane, are situated, two and two, on opposite sides of the 
plane, and at equal distances from it. 

97. Prop. The center of gravity of every homogeneous body 
of uniform density, symmetrical with respect to a plane, is situ- 
ated in that plane. 

For any two particles, symmetrically placed, will be at the 




46 



STATICS. 



same distance from the plane, and their moments will be equal 
and have contrary signs. But all the particles, taken two and 
two, are thus placed (Art. 96). Therefore the resultant of the 
system of forces will be in that plane, and, consequently, the 
center of gravity also. 

98. Def. A body is said to be symmetrical with respect to an 
axis when it is symmetrical with respect to two planes pass- 
ing through that axis. 

99. Prop. The center of gravity of a homogeneous body, sym- 
metrical with respect to an axis, is situated in that axis. 

By Art. 97, it must be in each plane passing through the 
axis, and therefore in their common intersection, or the axis 
itself. 

100. Cor. If a body is symmetrical with respect to two 
axes, its center of gravity will be at their intersection, since it 
must be in both axes. 

101. Def. This point is also called the center of figure. 

102. Prop. To find the center of gravity of any number of 
heavy particles whose weights and positions are given. 

Let A, B, C, &c, be the particles, whose weights w 1 , w 2 , 

w 3 , &c, act at their respective 
centers of gravity, vertically 
downward, and therefore con- 
stitute a system - of parallel for- 
ces. m)j and w 2 have a result- 
ant =w l +w 2 acting at some 
7 point a, such that w l .Aa=w 2 . 
Ba. The distance AB being 
given, the distanceA a is de- 
termined by taking (Art. 39) 

AB : A«=AB.- 




m 



W3 



w 1 +w 2 : w. 



w 1 -\-w. 



Compounding the weight w 1 +w 2 at a with another weight 
w 3 acting at C, they will have a resultant w 1 +w 2 +w 3 acting 
at some point b, such that 

(w 1 -hw 2 ).ab=w 3 .Cb. 



CENTER OF GRAVITY. 47 

First determining aC from the triangle aCB, in which BC, 
B«, and the angle aBC are supposed to be known, we can de- 
termine the distance ab by the proportion 

w l +w 2 +w 3 : w 3 =Ca : ab=Ca.- 



~i >w 2 +w 3 

By continuing the same process, we should determine the 
point at which the final resultant weight acts. The point will 
be the same, whatever be the order in which we compound the 
weights. 

103. Prop. To find the center of gravity of any number of 
particles in the same plane whose positions are given by their 
co-ordinates. 

Since the weights of the particles constitute a system of par- 
allel forces, let P a , P 2 , P 3 , &c, represent the weights of the 
particles which may be supposed collected in their respective 
centers of gravity, x x ,y x ,x 2 ,y 2 , &c, their co-ordinates. We 
shall have for the co-ordinates x, y of the center of parallel for- 
ces (Art. 44), or center of gravity of the whole body (Art. 94), 

__ P 1 ^ 1 +P 2 ^ 2 +P 3 3:3+,&c. _ S.Px "| 
* P 1+ P 2 +P3+,&c. -2.P I 

- P,yi+P,y,+P3ya+»&c. _s.p y f W 

y ~ P a +P 2 +P3+,&c. "LP'J 

104. Cor. 1. If the particles all lie in a straight line, this line 
may be taken for the axis of x, and y 1 =0, y 2 =0, &c. .*. y=Q, 
and the center of gravity will be in that line. 

105. Cor. 2. If the particles are homogeneous, the weights 
of each particle (24) will be proportional to the volumes ; and 
if «j, v 2 , v 3 , &c, denote the volumes of the particles, and V 
the whole volume, we have 

__v 1 x l -\-v 2 x 2 +v 3 x 3 +, &c._ 2.vx ^ 
x ~ V ~~^v~ ! 

y= v = sTj 

Hence the sum of all the particles, or the whole volume, multi- 
plied by the distance of its center of gravity from a plane, is 



48 STATICS. 

equal to the sum of each particle into the distance of its center 
of gravity from the plane. 

106. Cor. 3. If the center of gravity of the whole volume 
be given, and the center of gravity of one of its parts, the cen- 
ter of gravity of the other is readily obtained. 

Let V equal the whole volume, and v 1? v 2 the volumes of 
the two parts, x,y,x l ,y 1 ,x 2 ,y 2 , the co-ordinates of their re- 
spective centers of gravity. 

Then x— ' y > and U= y • 

But, by hypothesis, V, v 1 and x,y,x l ,y 1 are given. 
Therefore, *> 2 = V — v 19 

XX — V X VV — 15 1/ 

and x 2 = — L , and L=— l ^ 1 . (30) 

v 2 *- v 2 v ' 

107. We shall now proceed to apply the foregoing princi- 
ples to specific cases. 

Ex. 1. To find the center of gravity of a uniform physical 
straight line. 

If AB be the uniform straight line, and C its middle point, C 
r will be its center of gravity ; 

1 for we may consider the line 

made up of a series of equal particles in pairs on opposite 
sides of C, and the weights of each pair would be equal paral- 
lel forces having their resultant at C, the middle point between 
them, or the resultant of all the forces will pass through C. 

Or, the line AB is symmetrical with respect to a plane pass- 
ing through C. The center of gravity, therefore (Art. 97), is in 
C this plane, and as it is also in the line 
AB, it must be at their intersection C. 

Ex. 2. To find the center of gravity 
of a thin triangular plate of uniform 
density. 

Let ABC be the triangular plate, 
of which the thickness is inconsider- 
able. Bisect AB in D, and AC in E. 
^~~ -"]} Join C, D, and B, E intersecting in 

G. G will be the center of gravity of the plate. 




CENTER OF GRAVITY. 



49 



K 



Since the line CD bisects all lines drawn parallel to the 
base, and, consequently, divides the triangle symmetrically, 
the center of gravity of the triangle will be in this line. For 
the same reason, it will be in the line BE, and will therefore 
be at their intersection G. 

Join D, E. DE is parallel to BC, since it divides the sides 
AB and AC proportionally, and DE=|BC. 

From the similar triangles, DEG and BCG, we have 
DE : BC=DG : GC=1 : 2. .-. 2DG=GC. 
Adding DG to both sides, 

3DG=DG+GC=DC. .-. DG=iDC. 

In the same way it may be shown that EG=JEB. 

Hence the center of gravity of a triangle is one third the 
distance from the middle of either side to the opposite vertex. 

Ex. 3. To find the center of gravity of a parallelogram of 
uniform density and thickness. 

Bisect the sides AB and DC in H and K ; also the sides 
AD and BC in E and F. The 
plane passing through H and K 
will divide the parallelogram sym- 
metrically, since it will bisect all 
lines parallel to AB. The center 
of gravity will lie in this plane, and 
will therefore lie in its intersection HK with the parallelogram. 
For the same reason it will lie in EF, and must therefore be at 
G, the common point of these lines. 

Or, since each diagonal bi- 
sects all lines drawn parallel to 
the other, it will be at the inter- 
section of the diagonals. 

Ex. 4. To find the center of 
gravity of a thin, polygonal 
plate, of uniform density and 
thickness. 

Let ABCDEF be the poly- 
gon. Draw the lines AC, AD, 
AE, dividing it into triangles. 

D 



H 




50 



STATICS. 



When the polygon is given, these triangles will be known, 
and their centers of gravity, g lt g 2 > g 3 > g^ ma y be found by 
Ex.2. 

The mass of each triangle may be considered as a heavy 
particle at its center of gravity, or the weight of each, as a 
force acting at its center of gravity. Then {Art. 39), 

gi+g 2 '-g 2 =glg 2 : gl G, =glg2'~r^-' 

This determines the point G', the center of gravity of the 
portion ABCD. In the same manner, we may find the point 
of application G", of the resultant of g l +g 2 acting at G', and 
g 3 acting at g 3 , and so on ; the last point so determined will 
be the center of gravity G of the polygon. 

Ex. 5. To find the center of gravity of a triangular 'pyramid 
of uniform density. 

Let ABCD be the triangular pyramid. Bisect the edge BC 
in E, and pass a plane through E and the edge AD. This 




plane will bisect all lines in the pyramid parallel to BC, and 
will therefore divide it symmetrically. Bisect AC in F, and 
the plane through F and the edge BD will also divide the pyr- 
amid symmetrically ; and since the center of gravity of the 
pyramid will be in both these planes, it must be in their inter- 
section D/. But the point/ is the center of gravity of the face 
ABC {Ex. 2). Hence the center of gravity of the pyramid 



CENTER OF GRAVITY. 



51 



lies in the line drawn from a vertex to the center of gravity of 
the opposite face. Take Ee=JED, and join Ae. Since e is 
the center of gravity of the face BCD, the center of gravity of 
the pyramid will be in the line Ae. It must therefore be at 
the intersection G of Ae and Df. 

To find fG, join fe, which will be parallel to AD, since it 
divides the sides ED and EA of the triangle AED proportion- 
ally. Now the similar triangles fGe and AGD, with/Ee and 
AED, give 

/G : GD=fe : AD=Ee : ED=1 : 3. 
.-. 3/G=GD. 
Adding /G to both members, 

4/G=:/G+GD= /D, 
fG=-ifD. 

Hence, the center of gravity of a triangular pyramid is one 
fourth the distance from the center of gravity of one face to the 
opposite vertex. 

Ex. 6. To find the center of gravity of a pyramid whose base 
is any polygon. 

In the pyramid A BCDEF find G, the center of gravity of 
the polygonal base (Ex. 4), 
and join AG. Since AG 
passes through the center 
of gravity of the base, it 
will pass through the cen- 
ter of gravity of every sec- 
tion parallel to the base, 
and the center of gravity 
of the whole pyramid will 
be in AG. 

Join GB, GC, GD, GE, i 
and GF, and conceive 
planes to pass through A 
and each of these lines, 
thus dividing the whole 
pyramid into as many triangular pyramids as the base has 




52 STATICS. 

sides. The centers of gravity of these pyramids will be at 
one fourth the distances, respectively, from the centers of grav- 
ity of their triangular bases to the common vertex A. These 
distances being thus divided proportionally, the points of di- 
vision will all lie in the same plane parallel to the base. And 
since the centers of gravity of all the triangular pyramids are 
in this plane, the center of gravity of the whole pyramid will 
be in it, and, being in the line AG also, will be at their inter- 
section g. But the plane divides all lines drawn from the ver- 
tex A to the base proportionally ; therefore, the center of grav- 
ity g of the whole pyramid is one fourth the distance from the 
center of gravity of the base to the vertex. 

Cor. Since the above principle is true, whatever be the num- 
ber of sides of the polygon, it is true when the number becomes 
indefinitely great, or when the base becomes a continued closed 
curve, as a circle, an ellipse, &c. ; or, the center of gravity of a 
cone, right or oblique, and on any base, is one fourth the dis- 
tance from the center of gravity of the base to the vertex. 

Ex. 7. To find the center of gravity of a frustum of a cone or 
pyramid cut off by a plane parallel to the base. 

Let a be the length of the line drawn from the vertex of the 
cone, when complete, to the center of gravity of the base, a' 
that portion of it between the vertex and the smaller base of 
the frustum. Then (30) we have 

_ vx— v x x x 



in which x=\a, x^=\d. Now the part of the cone or pyra- 
mid cut off is similar to the whole, and similar solids are as the 
cubes of their homologous dimensions, or cubes of their lines 
similarly situated. 

Hence v : v^=a 3 : a' 3 , or v— v l : v=a 3 —a' 3 : a 3 . 



V' 



and 



v 2 =v-v^v(l-^). 



CENTER OF GRAVITY. 



53 



v.%a- 



<i>' 



a —a' 



•0-9 



./»» 



(a+a')(a 2 +a n ) 

tf+aa'+a' 2 ' 



Subtracting this from a, we have the distance of the center 
of gravity of the frustum from the center of gravity of its base 
equal to 



(a+a')(a' + a'*) a 



Sa' 3 



aa'+a' 2 4 ±(a 2 +aa'+a 12 )' 




Ex. 8. To find the center of gravity of the perimeter of a tri- 
angle in terms of the co-ordinates y 
of the angular points. 

In the triangle ABC, let a, &, 
c represent the sides respective- 
ly opposite to the angles A, B, 
C. Their centers of gravity will 
be each at the middle point of 
the side; as g l9 g 2 ,g 3 . ° x 

Let x l y 1 be the co-ordinates of A referred to the origin O, 
x 2 y 2 " " " B, 

x 3 y 3 « " " C. 

Then the co-ordinates of g A are \{x^-\-x 2 ), iO/j+f/s)* 

g 3 " i(x 3 +x 1 ),i(?/3+y 1 ), 
and x, y, being the co-ordinates required, by (28), we have 
__ a(x +x 3 )+b(x 1 +x 3 )+c(x l +x 2 ) 
X ~~ ~2(a+b+c) 

a(y 2 +y 3 )+Hy,+y3)+ c (yi+y 2 ) 

2(a+fc+c) 

Ex. 9. To find the co-ordinates of the center of gravity of a 
triangle. 

Letx J y 1 , x 2 y 2i x 3 y 3 be the co-ordinates of the points A. 



y- 



54 



STATICS. 




O L 



3Sf M 



B, C respectively. Draw 
AD bisecting BC in D, 
and take AG=f AD : G 
is the center of gravity 
of the triangle. The co- 
ordinates of D are \{x 2 

+^ 3 )>i(2/2+3/ 3 ); and if 
£=ON, y=GN, be the 

co-ordinates of G, we 

have 



ON=OL+f(OM-OL), 
GN=AL+f (DM-AL) ; 
or x=x x +%{U X 2 +*»)''— *-i } =i( x i+ x 2 +* 3 )> 

and y=yi+t{i(y a +y 3 )-yi}=i(yi+y a +ys)- 



CONDITIONS OF EQUILIBRIUM OF BODIES FROM THE ACTION OF 

GRAVITY. 

108. Prop. If a body have a fixed point in it, the condition of 
equilibrium requires that the vertical line through the center of 
gravity shall pass through the fixed point. 

If the center of gravity g be in the vertical line Ag, passing 

through the fixed point 
A, the weight w of the 
body, being a vertical 
force acting at g, in the 
direction Ag, will be 
resisted by the reaction 
of the fixed point A. 
If the center of gravity 
be at any other point 
g', then drawing the 
vertical line through g 1 
and the horizontal line through A, the weight w acting at g' 
would have an uncompensated moment w.Am, which will not 
vanish until the center of gravity comes into the line Ag. Or, 
if g' be the center of gravity, the body will be acted upon by 





CENTER OF GRAVITY. 



55 



a couple of which the forces are, the weight of the body at g', 
and the reaction of the fixed point A,, and the arm Km. 
Therefore (19), in order to equilibrium, 
2(X.y—Yx)=w~Am=0. 

.'. K?n = 0, or the point g' must be in Kg. 

109. Def. The equilibrium is said to be stable when the body, 
if slightly disturbed, tends to return to its original position. 
It is called unstable when, being disturbed, it tends to re- 
move further from its original position ; and neutral when, after 
being disturbed, it still remains in equilibrium. 

110. Prop. When the equilibrium of a body containing a 
fixed point is stable, the center of gravity is in the lowest posi- 
tion it can take ; when unstable, in the highest. 

Let A be a fixed point in the bodies M and N, g their cen- 
ter of gravity. The 
center of gravity can 
only move on the 
surface of a sphere 
whose center is A 
and radius Kg. When 
the body M is dis- 
turbed, its center of 
gravity g being re- 
moved to g', will rise 
through the versed sine of the arc gg', and when at g', the 
moment iv.Km will obviously tend to bring it back to its orig- 
inal position. The equilibrium is therefore stable, and the 
center of gravity the lowest possible. In the body N, the cen- 
ter of gravity being at g is the highest possible, and being in 
the vertical Kg, will be in equilibrium. When removed to g', 
the moment w.Km will obviously tend to carry it further, and 
the equilibrium was therefore unstable. 

111. Cor. 1. The pressure on the point by which a body is 
suspended is clearly, in the case of equilibrium, equal to the 
weight of the body. 

112. Cor. 2. If a body is suspended from two points, the po- 
sition of equilibrium is that in which the center of gravity is in 




56 



STATICS. 




the vertical plane passing through the two points of suspen- 
sion, since it is then the highest or low- 
est possible. To determine the pres- 
sures on the fixed points A and B, let 
GC represent the weight of the body- 
acting vertically at G. Resolve GC 
into the two forces DG and EG acting 
in the directions AG and BG. These 
will represent the pressures on A and 
B. Since the directions of GC, GD, 

and GE, and the magnitude of GC are known, the magnitudes 

of DG and EG may be determined. 

113. Cor. 3. If a body be suspended from three fixed points 
not in a right line, the body is necessarily at rest. 

With regard to the pressures on each point, the three lines 
drawn from the fixed points to the center of gravity give the 
directions of the pressures, and the vertical is the direction of 
the weight, the magnitude of which is given. Hence we have, 
in a parallelopiped, the three sides and diagonal given in po- 
sition and one given in magnitude, to determine the magnitude 
of the other three. 

114. Cor. 4. If a body touch a horizontal plane in one point, 
it will be in equilibrium when the vertical through its center 
of gravity, and the perpendicular to the plane at the point of 
contact, coincide ; for the weight will then be counteracted by 
the plane. 

115. Cor. 5. If the body touch the plane in two points, it 
will be in equilibrium when the verti- 
cal through the center of gravity, and 
the perpendiculars to the plane at the 
points, are in the same plane. Thus, 
if ABC be a vertical section through 
the two points of support P and Q, 
there will not be an equilibrium unless 
the vertical through the center of grav- 
ity G is in the same plane, or, which is the same thing, meets 




CENTER OF GRAVITY. 57 

the line PQ. The pressures on P and Q may be determined 
by the theory of parallel forces, and 

P : Q : W=WQ : WP : PQ, 

P, Q, and W representing the pressures on the two points and 
the weight of the body respectively. 

If the body touch the line PQ in more than two points, the 
problem of the pressures is indeterminate, as the pressures 
may be any how distributed. 

116. Cor. 6. If the body touch the plane in three points, it 
will be in equilibrium when the vertical through the center of 
gravity falls within the triangle formed by joining these points. 

To estimate the pressures in this case, let PQR be the tri- 
angle formed by joining the three 
points, and W the point where the 
vertical through the center of 
gravity meets it. We must re- 
solve the weight acting at W into 
three others parallel to it, acting at 
P, Q, and R. Join PW, Q W, and 
RW, and produce them to the op- 
posite sides. Then, by the theory of parallel forces, 

P : W=WK : TK=triangle WRQ : triangle PRQ, 
Q : W=WM : QM= " WRP : " PRQ, 

and R : W=WH : RH= « WPQ: " PRQ. 

.-. P : Q : R : W= WRQ : WRP : WPQ : PQR. 

If the body touches the plane in more than three points, the 
pressures on the points are indeterminate, but their sum is 
equal to the weight of the body. 

117. Cor. 7. If the vertical through the center of gravity of 
a body on a plane meets the plane in a point within the base, 
the body will stand. For the resultant of the parallel forces 
of resistance must be within the figure formed by joining the 
several points of contact. 

If the vertical falls without the base, we have two parallel 
forces in contrary, but not opposite, directions, and the body 
will turn over. 




58 



STATICS. 




118. Prop. The stability of a body is measured by the excess 
of the shortest line that can be drawn from the center of gravity 
to the perimeter of the base, above the vertical, from the center to 

the horizontal plane. 

The stability will depend on the ex- 
cess of GP over GH, since G must be 
elevated a distance equal to this dif- 
ference, in order to turn the body over 
the edge of the base at P. 

Cor. 1. The greater the base HP the greater the stability, 
if the height of G remain the same ; and the greater HG, the 
less the stability if HP remain the same. 

Cor. 2. The stability is measured by the versed sine of the 
arc, through which the center of gravity must move from rest 
to its highest point. For GP-GH=GP-BP= AP-BP=AB 
== versed sine of arc GA to radius PG. 

119. Prop. If a body be placed on an inclined plane, it will 
descend when there is no resistance from friction. 

For the weight of the body, represented by GA, may be re- 
solved into two others, GB 
and BA, one perpendicular 
to the plane, and the other 
parallel to it, of which GB, 
the one perpendicular to the 
plane, can alone be counter- 
acted by the plane. In this 
case, if the vertical GA fall within the base of the body, the 
body will slide ; if it fall without, it will slide also. 




120. 



EXAMPLES. 



1. If two right cones have the same base and their vertices 
in the same direction, find the distance of the center of gravity 
of the solid contained between their two surfaces from their 
common base. 

Ans. \ sum of their altitudes. 



CENTEROFGRAVITY. 59 

2. The center of gravity of a paraboloid being in the axis 
at a distance from the vertex equal to f of the axis ; find the 
center of gravity of a frustum of a paraboloid from the base, 
a and b being the radii of the two ends, and m the parameter 
to the axis (30). 

a 6 -3a*b*+2b« 
AUS ' 3m(a 4 -6 4 ) * 

3. Two spheres, whose radii are a and b, touch each other 
internally ; find the distance of the center of gravity of the solid 
contained between the two surfaces from the point of contact. 

tf+cfb+aV+b 3 
a +ab + b 2 

4. The distance of the center of gravity of a hemisphere 
from its base being § the radius, find that of a hemispherical 
bowl whose internal radius is a and thickness c. 

a 4g 3 +6g 2 c+4flic 2 +c 3 
AnS ' ¥ * 3a 2 +3ac+c 2 ' 

5. From the result obtained in Ex. 4, find the distance of the 
center of gravity of a hemispherical surface from the center 
of the base. 

Ans. \a. 

APPLICATION OF THE PRINCIPLES OF THE INTEGRAL CALCULUS TO 
THE DETERMINATION OF THE CENTER OF GRAVITY. 

121. By the principles of the integral calculus, when the 
volumes v (Art. 105) become indefinitely small, they may be re- 
garded as the differential elements of the body, and be repre- 
sented by dv. In this case formulas (29) will take the form 

x -/dv' y~/d»' (31) 

in which x and y denote the distances of the center of gravity 
of dv from the co-ordinate axes. 

122. Prop. Required the differential expressions for the co-or- 
dinates of the center of gravity of a plane curve or line. 

If ds represent the differential element of the curve or line, 
by substituting ds for dv in (31), we have 



60 STATICS. 

_y^ _j^ (32) 

If the arc is symmetrical with respect to the axis of x, the 
center of gravity will be in that axis (Art. 99), and y=0. 

_ fxds 
.'. x= % — - 
s 

is sufficient. 

123. Prop. Required the differential expressions for the co- 
ordinates of the center of gravity of a plane area. 

Since the differential element of a plane area is dxdy, dv= 
dxdy. By substitution in (31), we have 

- = ffxdxdy - = ffydxdy 
ffdxdy ' y ffdxdy ' 
Integrating in reference to y, we have 
_ Jxydx fjfdx ^ 

fydx ' y fydx ' y ' 

If the area is symmetrical with respect to the axis of x, the 
center of gravity is in that line (Art. 99), and y=0. 

__fxydx 

~ fydx 

is sufficient. 

124. Prop. Required the differential expressions for the co- 
ordinates of the center of gravity of a surface of revolution 
around the axis of x. 

The center of gravity will obviously be in the axis of x, and 
therefore y=0; and since, for a surface of revolution, dv= 
2nyds, the first of equations (31) become, 

5-^- (34) 

X /yds' (di) 

and this equation is sufficient. 

125. Prop. Required the differential expressions for the co* 
ordinates of the center of gravity of a solid of revolution. 

In this case dv=ny 2 dx. Hence, from the first of equations 
(31), we have 



CENTEROFGRAVITY. 61 

fv^xdx 

which alone is sufficient. 

By proper substitutions for do in the fundamental equations 
(31), we may find expressions for the co-ordinates of the cen- 
ter of gravity for other forms of bodies. 

126. Prop. The surface generated by the revolution of a curve 
around an axis is equal to the length of the curve, multiplied by 
the circumference described by its center of gravity. 

From the second of equations (32), we have 

2n.y.s=2nfyds. 

Now 2v:y is the circumference of which y is the radius, and 
2ny.s is the circumference described by the center of gravity 
of the curve s in its revolution round the axis of x, multiplied 
by the length of the curve 5. But this is equal to 2nfyds, which 
is the area of the surface generated by the revolution of the 
curve. Hence, &c, 

127. Prop. The volume generated by the revolution of a plane 
area around an axis is equal to the product of that area by the 
circumference described by its center of gravity. 

For, from the second of equations (33), we have 

2nyfydx = nfy 2 dx. 

In this equation, fydx is the generating area, 2ny is the cir- 
cumference described by its center of gravity, and t nfy 1 dx is 
the volume generated. Hence the truth of the proposition. 

These last two propositions comprise the theorem ofGuldin, 
and their application to the determination of the surfaces and 
volumes of bodies constitutes the Centrobaryc Method. By 
this method, of the three quantities, viz., the generatrix, the dis- 
tance of the center of gravity from the axis, and the magnitude 
generated, any two being given, the other may be determined. 



62 



STATICS. 



128. EXAMPLES. 

1. Required the center of gravity of a circular arc. 

Let the axis of x bisect the arc MAM' 

in A, the origin being at the center of 

the circle, and let MAM' = 2s. From the 

x equation of the circle ?/ 2 =r 2 — x 2 , we ob- 

. dy*_ z* 




tain 



dx" 



But ds=dx\J l+-p a = 



rdx 



,. (32) **£** 



vV- 

/xdx 



ry 



vV 



+-Vr*-x*+C=+-Z+C. 
s s 



TV 

When y=0, x=r, and —■ 
Hence 



r. 



.\ c=o. 



s 2s 

or, the distance of the center of gravity of a circular arc from 
the center of the circle is a fourth proportional to the arc, the 
radius, and the cord of the arc. 

If the arc be a semicircle, y—r, and s=\m\ 



\ CI= 



2r 



0.63662r. 



2. Required the center of gravity of a circular segment. 
Putting CP=a (Fig., Ex. 1), and taking the center for the 

~. Hence (33), 



origin, we have y= Vr 

flxydi 
fydx 



__f; xijdx _ f;x(r*-xy-dx _ i(r*-ay _±MP 
MAP " MAP "MAP' 
V(2MP) 3 T \ (chord) 3 



* * IMAM'P segment 
If the segment is a semicircle, 
- T V(2r) 3 4r 



*7rr 



3rr 



0.42441r=fCI. 



CENTER OF GRAVITY. 63 

3. Required the center of gravity of the surface of a spher- 
ical segment. 

Taking the origin at C (Fig., Ex. 1), the center of the gen- 

dy 1 x* 
erating circle, we have z 2 +?/ 2 =r 2 , — =-^, and yds—rdx. 

... (34) -J'yxdsJ-xdx^ 

v ' /yds fdx ' 

Integrating between x=r and x=a=CF, 

x=^ 7 =i(r+a). 

r—a * v ' 

Hence the center of gravity is at the middle of PA. 

4. Required the center of gravity of a spherical segment. 
Taking the origin at A, the vertex of the generating circle, 

we have, for its equation, y' 2 =2ax—x 2 . 

Y 

fxy*dx _f (2ax—x*) xdz 
.'. (35) x ~f y * dx = f(2ax-x*)dx ' 

lVN Ux 3 -\x* 8ax-3x* 
HenCe AG= ^^P^ = T2^4x' 

If the segment is a hemisphere, x=a, 
and x=^a. 

5. Required the surface of a hemisphere. 

By the centrobaryc method (Art. 126), we have 

The generatrix =^nr, the ordinate of its center of gravity 

2r 
y=— (Art. 128, Ex. 1), and the circumference described by 

2r 

the center of gravity =2n.— = 4r. Hence the surface =^7rrAr 




■2irr\ 



129. EXAMPLES ON THE PRECEDING CHAPTERS. 



Ex. 1. Two beams, rigidly connected at a given angle, turn 
on a horizontal axis through their point of union ; find the po- 
sition of equilibrium by the action of their own weights. 

Let AC, BC be the beams suspended from C, and making 



64 



STATICS. 




with each oth- 
er the angle a. 
Since C is a 
fixed point, the 
only condition 
of equilibrium 
is, that the sum 
of the moments 
about C is zero (Art. 75). 

Let g lt g 2 be the centers of gravity of the beams, and g^C 
=a,g 2 C = b. Also, w 1 = weight of AC, acting at g-^and w 2 = 
weight of BC, acting at g 2 . Draw through C the line MCN 
horizontally, meeting the vertical directions in which w 1 and 
w 2 act, at M and N. 

By (19), w 1 CM-ii; a CN=0. Let BCN=0. The determin- 
ation of will fix the position of the compound beam. 

Since CM=^C. cos. MCA and CN=g- a C.cos. BCN, we 
have 

w 1 .Cg l cos. ACM-u> 2 .Cg- 2 cos.BCN=0, 
w 



or 



a. cos. (180 — a+d) — w 2 .b. cos. = 0. 
.-. Tan. 6- 



w 2 b+w l a. cos. a 



Tan. 6- 



w x a. sin. a 

Ex. 2. When a given weight W is hung from the end of 

one of the beams, A (Ex. 1), find in case of equilibrium. 

w 2 .b + (w 1 +2W)<2 cos. a 

(w J +2W)asm. a 

Ex. 3. Two beams, as in Ex. 1, are suspended from one 

end B ; find the angle which the upper one makes with a 

horizontal line. 

m „ (2w 1 +w 9 )b— w,a. cos. a 
Tan. 0=- — l 2 - — r— ± . 



N. B. Since the common center 
of gravity of the two beams is in the 
vertical through B, BM=Gm=C7?* 
-CG=Cm-2NB=a. cos. (a-6)- 
2b cos. 0. 




EXAMPLES ON THE PRECEDING CHAPTE 



65 



.*. w 2 b. cos. 6—w 1 a. cos. (a— 6)-\-2w 1 b. cos. 6=0. 

Ex. 4. Two spheres of unequal radii, but of the same ma- 
terial, are placed in a hemispherical bowl ; find the position 
they take when in equilibrium. 

Since the reactions of the bowl upon the spheres are in the 
directions of the radii of the 
spheres through the points 
of contact, and since these 
radii produced pass through 
C ; if C was a fixed point, 
and connected with A and 
B by a rigid rod without 
weight, the bowl might be 
removed without disturbing 
the equilibrium. The ques- 
tion, then, is reduced to finding the position of equilibrium of 
two weights suspended from the extremities of two rigid rods 
without weight, and is solved like the preceding. This posi- 
tion will be known when 6 is known. 

Let R be the radius of the bowl, r x , r 2 the radii of the 
spheres A and B respectively, and ACB=#. 

Then AB=r 1 -j r r 2 , CA^R-r,, CB=R-r 2 , 
AC 2 +CB 2 -AB 2 



M 


c 


1ST 




~r 




J6 




\ ( A 


/_ 


J\ 




B \y 


y 


f 




V 



and cos. a- 



2AC.BC ' 
(R- ri y + (R-r 2 y-( ri +r 2 y 



, which gives a. 



2(R-r 1 )(R-r 2 ) 

Then (19) mj 1 .CM-^ 2 CN=0, or rJ.CM-rJ.CN^O ; since 
the weights of the spheres are as the cubes of their radii. 
Substituting the values of CM=(R-r 1 ) cos. (180- (a+0) ), 
and CN=(R— r 2 ) cos. 0, expanding and reducing, we get 

Tan. e= r32(R ~ r2)+r;(R ~ ri) cos - a . 
r\(R— rj sin. a 

Ex. 5. A heavy beam rests upon a smooth peg with one 
end against a smooth, vertical wall ; find the position of equi- 
librium. 

E 



66 



STATICS. 




Let ACB be the beam, resting at A against the wall ADE 

and upon the peg C. 

The center of gravity g, 
' when there is an equilibrium, 
will evidently be at some 
point beyond C from A. Let 
Kg— a, DC=o, w= the weight 
of the beam acting at g f R= 
the reaction of the wall per- 
pendicular to itself at A, and 
R'= the reaction of the peg 
perpendicular to the beam at 
C. The angle 0, which the beam makes with the horizontal 
direction when in equilibrium by the action of these three 
forces, is required. 

Employing (17), (18), and (19), and resolving the forces in 
vertical and horizontal directions, and about the point C, we 
have 

w resolved in a horizontal direction = 0, 
R " " " 

R' " " " 

..-. (17) 2.X = R-R'sin. 0=0. 
Also, -to estimated vertically 
R 
R' 

.-. (18) 2.Y=w-R'cos.0=O. (b) 

Also, the moment of to about Q = w.C¥=w.(DY— DC). 
But DF=AH = Ag-cos.0=acos.0, and DC=o. 
.-. w.CF = w.(a. cos. 0— b), 

the moment of R about C=R.CK=R.AK. tan. 0=R.&. tan. 5 

R' " =0. 
,\ (19) 2.(Xy-Yx) = w.(a.cos.d-by-R.b.ten.6=0. (c) 
Multiplying (a) by cos. 6 and (b) by sin. 0, and subtracting, 



= 


R, 






— 


-R' 


sin. 


0. 
(a) 


= 


w, 






= 


0, 






= 


-R' 


cos 


. e. 



we have 



R. cos. 0— w sin. 0=0 
.-. R = totan. 0. 



EXAMPLES ON THE PRECEDING CHAPTERS. 67 

Substituting this value of R in (c), 

w(a cos.6— b)—w.b. tan. 2 0=0, 

or a cos. 0=6(1 + tan. 2 0) = b. sec. 2 0= b. j-r. 

v 7 cos. 

3 /& 
.*. cos. 0= \/ -, and &<«, except when 0=0. 

Ex. 6. Solve .E:c. 5 by resolving the forces parallel and per- 
pendicular to the beam, and taking the moments about either 
A or g. 

The three forces resolved in the direction of the beam give 
R cos. 0— w sin. 0=0; (a) 

resolved perpendicularly to the beam, give 

R'-io cos. 0-R sin. 0=0. (b) 

The moments about A give 

R'.b. sec. 0— w.a. cos. 0=0. (c) 

From (a), we have R=w tan. 0, which, substituted in (6), 
gives 

(cos. 2 0+ sin. 2 0) 
R' = w (cos. 0+ tan. sin. 0)=w ^-^ ' =u> sec. 0. 

This value of R', substituted in (c), gives 
w6 sec. 2 6—wa cos. = 0. 

.-. cos. 0= \/ -, as before. 
v « 

The moments taken about g- give 

R / .Co--Rasin.0=O, 
orR'(«— 6see.0) — R.«sin.0=R / «cos.0— R'b— Rasin.0cos.0=O. 

Substituting the values of R and R' above, and reducing, 



cos 



■e={/l 

V a 



Ex. 7. A heavy beam lies partly in a smooth hemispherical 
bowl and partly over the edge ; find the position of equilibrium. 

The beam ABC will be supported by the reaction R of the 
bowl at A, perpendicular to the surface, or in the direction of 
the radius AO, by the reaction R' of the edge of the bowl at 



68 



STATICS. 




B perpendicular to the 
beam, and by its weight 
w acting at g. 

Let Ag=a, AO=r= 
radius of the bowl, and 
0=ABO=BAO=incli- 
nation of the beam to 
the horizon. 

If the object be to 
determine the angle 
solely, it will be most 
readily effected by resolving the forces in the direction of the 
beam and taking the moments about B, by which we avoid 
expressions involving the unknown reaction R'. 

The reaction R in the direction AB= R cos. 0, 
the weight w " " =— w sin. 6. 

.*. R cos. d—w sin. 0=0, 
or R = w tan. 6. 

For moments about B, we have 

R.AB sin. 6— w.gB cos. 0=0. 
But AB=2r cos. 0, and gB=(2r cos. 0— a). 

.*. 2.R.r. sin. cos. 6—w(2r cos. 0— a) cos. 0= 
Substituting the value of R, and reducing, 

2r tan. 0. sin. 0— 2r cos. 0+a=O> 
or 2r— 4r cos. 2 6-\-a cos. 0=0. 



(a) 



<&) 



0. 



COS. = 



a± \/32r 2 +a 2 



8r 
in which the + sign only is admissible. 

Ex. 8. Solve the last example by following, step by step, 
the method of Art. 71, taking A for the origin of co-ordinates, 
and AB for the axis of x. 

Ex. 9. Find the horizontal strain on the hinges of a given 
door, and show that the vertical pressures are indeterminate. 

Let the annexed figure represent the door, of which A and 
B are the hinges. Let g be its center of gravity at which 
the weight w acts. The door is kept in equilibrium by the 



EXAMPLES ON THE PRECEDING CHAPTERS. 69 



weight acting at g, and the reac- 
tions of the hinges represented by 
the oblique arrows at A and B. 

Let A be the origin of co-or- 
dinates, AX the axis of x, AY the 
axis of y ; and let x=a, y=b be 
the co-ordinates of g ; x = Q, y=h 
those of the hinge B. Let the 
resolved parts of the reactions at 
B be Q,j horizontally, and R x vert- 
ically, and Q 2 , R 2 those at A re- 
spectively. 



% 



NT 



R2 



Ri 



a! % 



Then 2. X = Q 2 -Q 1= 0, or Q 1 =Q 2 , 
2. Y =w-R 1 -R 2 =0, 

2.{Xy-Yx) = w.a-Q, 1 h=0. 

From (a) and (c) we have Q 1 =-7-=Q, i , 



(P) 



which gives the horizontal strain ; and it is the same in mag- 
nitude at each hinge, but opposite in direction. 

Again, from (b) we have R l -\-R 2 =w, but we have nb oth- 
er relation by which we may determine the values of R a and 
R 2 , which are therefore indeterminate. 

Ex. 10. Two given 
smooth spheres rest 
in contact on two 
smooth planes, inclin- 
ed at given angles to 
the horizon ; find their 
position of equilibri- 
um. 

Let the planes AB, AC make the angles a and respective- 
ly with the horizontal line through A; O l and 2 be the cen- 
ters of the spheres at which their weights w x ,w 2 respectively 
act; Rj, R 2 the reactions of the planes at the points of con- 




70 



STATICS. 



tact, perpendicular to themselves, and therefore passing through 
the centers of the spheres to which they are tangents. Let S 
be equal to the mutual pressure of the spheres at their point 
of contact, acting in the line passing through their centers, and 
making the angle 6 with the horizontal line AD.d is required. 

Each sphere is in equilibrium from its own weight, the re- 
action of the plane against which it rests, and the pressure of 
the other sphere. 

By resolving the forces in the direction of each plane for the 
equilibrium of each sphere, we shall avoid equations involving 
the unknown reactions R, and R 2 , and have, in the direction 
of AB, 

w 2 sin. a— S cos. {a— 0)=O, (a) 

in the direction of AC, 



w 1 sin. 0-S cos. (0+0) = 0. 
Eliminating S, we have 

w 2 sin. a. cos. ((3+d) = w l sin. (5. cos. (a- 
Expanding and reducing, 
w sin. a. cos. (3. 



<P) 



6). 



tan. 6 



ifl, sin./J. cos. a 



io. 



cot. j3— 10, cot. a 



(w l -\-w 2 ) sin. a. sin. (3 w 1 +w 2 

Ex. 11. A sphere is sustained upon an inclined plane by the 
pressure of a beam movable about the lowest point of the in- 
clined plane ; given the inclination of the beam to the plane, 
required that of the plane to the horizon. 

Let AgB be the 
beam movable about 
A, 10= weight of 
beam acting at g, B 
the point of contact 
with the sphere 
whose center is C, 
w '= weight of the 
sphere. 

The sphere is in equilibrium, from the reaction R of the 
plane at the point of contact F, from the pressure P of the beam 
at B, and from its own weight w'. These three forces all act 
through the center C. Ag=a, AB = 6, BAD = a are given, or, 





<j 




C 




^SZeL 






E^ 


Vp 


^^ — i?r~ 










l 


> 


f w \ 


V 





D 



E 



EXAMPLES ON THE PRECEDING CHAPTERS. 71 



instead of either of the two latter, the radius of the sphere may 
be given. The angle DAE = 0, the inclination of the plane 
when in equilibrium, is required. 

For the condition of equilibrium of the beam, take the mo- 
ments about A. 

Y.AB=iv.Ag. cos. (a+d). 



Hence 



P=Y cos. (a+d). 



For the condition of equilibrium of the sphere, resolving the 
forces in the direction of AD, we have 

w' sin. 6— P sin. a=0, 



or 



whence 



w' sin. O—Wj sin. a cos. (a+d) ; 



tan. 6-- 



wa cos. a. sin. a 



wa sin. 2 a +w'b' 

which gives 6 the elevation of the plane, as required. 

Many statical problems require, for the determination of all 
the unknown quantities, equations to be formed by geometrical 
relations. Take, for illustration, the following : 

Ex. 12. A heavy beam turns about a hinge, and is kept in 
equilibrium by a cord attached to the lower end ; the cord 
passes over a pulley in the same horizontal line with the hinge, 
and sustains a given weight ; find the position of equilibrium 
of the beam. 

Let A be the hinge, C the 
pulley, AC=c, AB the beam 
= / in length, g its center of 
gravity at which its weight 
iv acts, Ag=a and P= the 
weight hung from the cord 
and measuring its tension t. 

Let d= ABC and 0=BAC, 
both unknown. 

Taking moments about A, we have, 

Z.AB. sin. 6=w.Ag cos. </>, 




72 



STATICS. 



or 



w a 
sin. 0=-.-. cos. 0. 

From the geometrical data, we have 
AC sin. 



(a) 



or 



AB 

sin. 6- 



sin. ACB' 

c 

i 



. sin. (0+0). 



<P) 



Equations (a) and (&) suffice to determine 6 and 0. 
.Ex. 13. A uniform beam rests with its lower end in a 
smooth, hemispherical bowl, and its upper end against a 
smooth, vertical plane. Find the position of equilibrium. 

The beam AB rests against the 
vertical plane AD at A and upon 
the bowl at B, and is sustained by 
its weight w acting at g, the reac- 
tion R of the bowl in the radius 
BC, and the reaction R' of the 
plane perpendicular to itself. 

Let r— radius of the bowl ; AB 
—2a, Ag=a, since the beam is 
uniform, w— weight, and CD =d, 
all supposed known. 

Resolving vertically, we have 




R sin. 6—w=0, or R= 



w 



sin. 6 



or 



Taking the moments about A, we have 

R.AB. sin. (d—(p)—w.kg. cos. 0=0, 
R.2<2. sin. (d—(p) — w.a. cos. <p=0. 
sin. (0—(f>) cos. <p 



" sin.0 2 °' 

or cos. 0— cot. 6. sin. §—\ cos. 0=0. 

Hence tan. 6=2 tan. 0. 

This equation containing two unknown quantities, a geo- 
metrical relation between them must be obtained. Cm being 
a vertical line meeting AB in m, 



(a) 



EXAMPLES ON THE PRECEDING CHAPTERS. 73 



cos. 6 Bra 2a— Am_2a— CD. sec.<f>_ 2a— dsec.<f> 
cos. <f)~ BC — r r r 



cos. 6= 



2a cos. cf)—d 



(») 



Equations («) and (fr) are sufficient to determine <j> and 0, as 
required. 

ifo. 14. A heavy beam has one end resting against a smooth 
vertical wall, and the other sustained by a cord, which is fast- 
ened at a point vertically above the point where the beam 
rests. Find all the forces which keep the beam in equilibrium. 

Let CB be the beam, AB the cord, A ^ 
and C the points on the vertical wall AD. 

Let w= weight of the beam,^ its center 
of gravity Cg—a, CB=Z, AB=c, and AC 
=h, all supposed known. The angles A, 
B, and C will be known. 

Let t= tension of the cord. The beam 
will press against the wall, and this press- 
ure may be resolved in a vertical and hor- 
izontal direction ; the latter, perpendicular 
to the wall, will be destroyed by its reac- 
tion ; but, since the wall is smooth, the 
vertical component can be balanced only 
by an opposite force =P. This vertical 
component, will be upward, or downward, or zero, according 
to the position of the point C. We shall suppose its position 
to be such that the component may be upward, and require a 
force to be applied downward to keep the end of the beam 
atC. 

Resolve vertically and horizontally, and take the moments 
about C. 

P+£cos. A— w=0, (a) 

R-*sin.A=0, (b) 

w.a. sin. C—t.L sin. B=0. (c) 

w.a. sin. C 




w.a.c 



From (c), t= 

I sin. B Lh 

Substituting the value of t in (6), 



= tension of cord. 



74 STATICS. 

CLC 

R=wjr. sin. A= pressure against the wall ; 
and from (a), 

V=w ( 1 — — cos. A J = force to be applied at C, 

to prevent the beam from sliding along the wall. 

Ex. 15. A weight w hangs from one end of a cord of which 
the other end is fastened to a vertical wall ; the cord is pushed 
from the wall by a rod tied to it, which is perpendicular to the 
wall. Find the pressure R of the rod on the wall when the 
cord makes the angle a with the wall. 

Ans. R = w. tan. a. 

Ex. 16. A heavy beam, AB = /, of which the weight is w, 
lies with the end A against a smooth vertical wall AD, and 
the end B on a smooth horizontal plane DB, making with it. 
the angle 6. The distance of its center of gravity g from B is 
a, and it is kept in equilibrium by a cord attached to it at B 
and fastened at D. Required the tension of the cord, the re- 
action R of the wall, and the pressure P on the plane. 

Ans. R = Z=w.y. cot. 6. V=w. 

Ex. 17. A body (weight =w) is suspended by a cord (length 
= 7) from the point A in a horizontal plane, and is thrust out 
of its vertical position by a rod without weight, acting from 
another point B in the horizontal plane, such that AB = ^, and 
making the angle with the plane. Find the tension t of the 
cord. 

I 

t=w- cot. 6. 
a 

Ex. 18. A triangular plate of uniform thickness and density 
is supported horizontally by a prop at each angle. Find the 
pressure on each prop. 

Ex. 19. A uniform beam rests on two planes inclined at an- 
gles a and f3 to the horizon. Find the inclination 6 of the beam 
to the horizon. 

sin. (j3— a) 



tan. = 



2 sin. a sin. (3' 



EXAMPLES ON THE PRECEDING CHAPTERS. 75 



Ex. 20. A uniform beam AB hangs by a string BC from a 
fixed point C, with its lower extremity A resting on a smooth, 
horizontal plane. Show that, when there is an equilibrium, CB 
must be vertical. 

Ex. 21. A uniform beam AB is placed with one end A in- 
side a smooth hemispherical bowl, with a point P resting on 
the edge of the bowl. If AB= 3 times the radius R, find AP. 

AP=1.838R. 

Ex. 22. A beam, whose weight is w and length 6 feet, rests 
on a vertical prop CD (= 3 feet) ; the lower end A is on a hor- 
izontal plane, and is prevented from sliding by a string DA 
(= 4 feet). Find the tension of the string. 

Tension =-r^zw. 

Ex. 23. A uniform beam, CB=2a, has one end C resting 
against a smooth, vertical wall AC, and the other sustained by 
a cord, whose length is c, fastened to the point A. Find AC 
when the beam is in equilibrium from its weight w, the tension 
t of the cord, and the reaction R of the wall. (Vide Ex. 14.) 

Ex. 24. Given the inclination i of the right line AC to the 
horizon and the weight of a heavy body W, to determine what 
force or weight P x acting in the given direction WM will be 
sufficient to sustain W on the line. 



The body W is kept upon the line by the action of the force 

Y 



P a in the direction 
WM, and its weight 
acting in the direction 
WE. The reaction of 
the line normal to its 
direction is a third 
force (Art. 76). 

Resolving P and W 
horizontally and vert- 
ically, we have 

2.X=P 1 cos. (e+i):fWcos.270 o = P 1 cos. (e+i), 
S.Y=P 1 sin. (e+i)+Wcos. 180° = P 1 sin. (e+i)-W. 




76 STATICS. 

Hence (20) becomes 

P, cos. (e+z) + (P 1 sin. (e+z)-W) tan. i=0; 

or, expanding and reducing, 

P a cos. e=Wsin. i, 

Wsin.i x 

or P,= . (a) 

1 cos. e v 

If the reaction of the line be required, resolve all the forces 
horizontally (8). This gives 

N. cos. (90° + i)+P 1 cos. (e+i)=-Q. 

... H^ w 

1 sm. z v ' 

or, by (a), N =w£M*2. (c) 

•^ v 7 cos. e v ' 

Iftr. 25. A given weight W is kept at rest on a circular arc 
by a weight P attached to a cord which pass- 
Y es over a point M in the vertical line MX 
w through C, the center of the circle. Required 

\ the position of W, supposing no friction at M. 
Resolving the forces P and W vertically and 
horizontally, we have, calling the angle PM W, e, 
2.X=W-Pcos.e, 
S.Y= -Psin.e. 
Let the co-ordinates of the point W be x, y, and call the dis- 
tance MC, a, and MW, I; then the equation referred to M as 
the origin, is 

(a- a;) a +y'=r», 
or y= Vr a — (a— x) 2 * 

Differentiating, we obtain 

dy a—x a—x 

dz~ Vr*-(a-x)~~ V 

x 

Substituting in (21), and recollecting that cos. e—y and 

V 
sin. e=y, we get 

w-p.f- P.f.JU W -p?-p^=o. 

I I ax I I 



EXAMPLES ON THE PRECEDING CHAPTERS. 77 



Hence 



1= 



aP 
W 



an equation which determines the distance of W from M 

M 
Ex. 26. Instead of a circle, as in Ex. 25, let the 
curve be a hyperbola with its transverse diameter m 
vertical, the point M being at its center. 




Ex. 27. Required a curve such that a given 
weight P, by a cord passing over a fixed point ' m 
without friction, will balance another given weight 
W at every point of it. 




;w 



Ex. 28. Required the co-ordinates of the center of gravity 
of a semiparabola whose equation is y*=px, height ==a, and 
base =b. 

Ans. x=%a, y=%b. 

Ex. 29, Required the center of gravity of the surface of a 
right cone. 

Ex. 30. Find the center of gravity of a paraboloid of revo- 
lution whose altitude is a. 

Ans, x=^a. 

Ex. 31. Find the center of gravity of a segment of a hyper- 
boloid whose altitude is a. 



CHAPTER VI. 

ON THE MECHANICAL POWERS. 

130. The general object of machinery is to transmit and to 
economize the action of certain forces at our disposal. The 
specific end is, sometimes to augment the action of which the 
power employed is capable when applied without the inter- 
vention of machinery ; sometimes, merely to change the direc- 
tion of the action ; and sometimes to regulate the velocity of 
the point to which the action is transmitted. 

The most simple machines are denominated Mechanical pow- 
ers, and are reducible to three classes, viz., the Lever, Cord, 
and Inclined Plane. 

The first class comprehends every machine consisting of a 
solid body capable of revolving on an axis, as the Wheel and 
Axle. 

The second class comprehends every machine in which 
force is transmitted by means of flexible threads, ropes, &c, 
and hence includes the Pulley. 

The third class comprehends every machine in which a 
hard surface inclined to the horizon is introduced, as the Wedge 
and the Screw. 

The force which is used to sustain or overcome any oppo- 
sition is called the Power ; the opposition to be overcome is 
called the Weight. This distinction in the names of the forces 
employed implies none in their nature. 

§ I. the lever. 

131. Def. A Lever is an inflexible rod capable of motion 
about a fixed point, called a fulcrum. The rod may be straight 
or any how bent. 

It is generally regarded, at first, as without weight, but its 



THE LEVER. 



79 



weight may obviously be considered as another force applied 
in a vertical direction at its center of gravity. 

Def. The arms of a lever are the portions of it intercepted 
between the power and fulcrum, and between the weight and 
fulcrum. 

132. Levers are divided into three kinds, according to the 
relative positions of the power, weight, and fulcrum. 

In a lever of the 
first kind, the fulcrum 
lies between the points 
at which the power 
and weight act. 



\ 



.k 



In a lever of the sec 
ond kind, the weight acts 
at a point between the 
fulcrum and the point of 
action of the power. 



In a lever of the third 
kind, the point of action of 
the power is between that 
of the weight and the ful- 
crum. 



B 



■w 



C 



133. Prop. Required the 
condition of equilibrium 
and pressure on the ful- 
crum when two parallel 
forces act on a straight 
lever. 



B 



r W 



80 



STATICS. 



Since the fulcrum C is a fixed point, by Art. 75, the sum of 
the moments of the forces about C must be zero. 

Aw Let a be the angle made by 

/ the direction of the forces with 

A. / ^ tne l ev er. From (19), we 

B C have 

W.BC. sin. a— P.AC. sin. a=o, 

the moment of P being nega- 
P tive, since it tends to produce 

motion in a direction opposite to that of W. 

Hence -p-=^, or, in case of equilibrium, the weight and 

power are reciprocally proportional to the distances at which 
they act from the fulcrum. ^ 

Hence (Art. 36) the resultant of P and W must pass through 
C, and the pressure on the fulcrum is equal to the algebraic 
sum of P and W, and acts in the direction of the greater. 

Cor. If the power equal the weight, the distances of their 
points of action from the fulcrum will be equal. 

For P=W gives AC=BC. 

134. Prop. Required the condition of equilibrium and the 
pressure on the fulcrum when any two forces in the same plane 
act on a straight lever. 

-q Let the forces P 

* "' y \\ and W make the an- 

/ \ \ gles a and (3 respect- 

/ \ \ ively with the lever, 

/ \ \ and let their direc- 

A a /\ „ X tions, when produced 

\ jCt { V r if necessary, meet in 

•^y \ . D. Since C is a fixed 
V ^"W" point in the lever, the 

R ^ sum of the moments 

of P and W about C must be zero. Hence 

W.CB. sin. ]3-P.AC. sin. a=0, 




THE LEVER. 



81 



or 



W 



AC. sin. a 



(2.) 




A 




CB. sin. (3' 

But AC. sin. a= per- 
pendicular from C on 
AD, and CB. sin. (3= 
perpendicular from C 
on BD. Hence the con- 
dition of equilibrium 
requires that the pow- 
er and weight should be 
inversely as the perpen- 
diculars from the ful- 
crum on their respective directions. 

Since the lever is in equilibrium by the actions of P and W 
and the reaction of the fulcrum, the resultant of P and W must 
be equal and opposite to that reaction. It will therefore pass 
through C, and be equal to the pressure on the fulcrum. 

To find R, we have {Art. 29) 

R 2 =P+W 2 +2PW. cos. ADB in Fig. (1), 
and R"=P a +W s +2PW cos. AD W in Fig. (2). 

But ADB= 180- (a+(3), and ADW=180-ADB=180- 
(P-a). 

.-. R 2 =P 2 +W 2 -2PWcos. (a+0), 
or R 2 =P 2 + W 2 -2PW cos. ((3-a). 

To find the angle 6 made by R with the lever, resolve par- 
allel and perpendicular to the lever, the reaction of the fulcrum 
being equal and opposite to R. Hence we have, in Fig, (1), 

P. cos. a— W cos. j3+R cos. 6=0, 
P sin. «+W sin. (3-R sin. 0=0. 



Tan. 6- 



P. sin. fl+W sin. (3 



in Fig. (8), 



W cos. (3— P cos. a ' 

P cos. a— W cos. (3— R cos. 6=0, 
P sin. a- W sin. 0-fR sin. 6=0. 

.-. Tan.0=: 



W sin. (3— P sin. a 



P cos. a- 
F 



■Wcos.jS* 



82 



ST Al IC 



]35. Cor. Whenever the lever is bent or curved, the condi- 
tion of equilibrium is the same. 




have 



or 



A p 




For, since the 








moments of the 


A 
M 






forces about the 
fulcrum in oppo- 
site directions must 






/ A 




/w 


be equal, in case 
of equilibrium, we 


\CM= 


=W.CN, 




W 


CM 






P = 


"CN' 





136. Prop. Required the condition of equilibrium and the 
pressure on the fulcrum, when any number of forces act in any 
direction in one plane on a lever of any form. 

Let Pj, P 2 , P 3 , &c, be the forces in one plane, 

Pm p 2 > Pv " tne distances of their points of applica- 
tion from the fulcrum, 
a l9 a 2 , a 3 , " the angles made by the directions of P 
and p respectively. 

Then p l sin.a 1 ,p 2 sin. a 2 , p 3 sin. a 3 are the perpendiculars 
from the fulcrum on the directions of the forces, and ¥ 1 p 1 sin. 
a,, &c, the moments of the forces. 

When there is an equilibrium, the sum of the moments of 
the forces about the fulcrum will be zero, or 

V 1 p 1 sin. tfj+Psfa sm - a 2+^3i°3 sin. <z 3 +,&c, =0. 

The signs of the moments will depend on the direction in 
which they tend to produce rotation. 

To find the pressure on the fulcrum, we must determine the 
resultant of all the forces supposed to concur at the fulcrum. 
For each moment 'P 1 p 1 sin. a^ is equal to a couple of which 
the forces are P and the arm p sin. a, and a single force P act- 
ing at the fulcrum. The magnitude of R will then be determ- 
ined by the equation (6), 

R= V(S.X)'+(S.Y) Q . 



THE LEVEE. 83 

Cor. If the lever be straight and the forces parallel, 

a l =a 2 =a 3 = , &c. 
Hence P 1 /> 1 +P 2 jt7 2 +P 3 ^3+, &c, =0. 

The same result will also be obtained by following, step by 
step, the method of Art. 71. 

EXAMPLES. 

Ex. 1. On a straight lever AB of the first kind, without 
weight, 36 inches in length, a weight W=15 lbs., acting at B, 
is balanced by a power P=3 lbs. acting at A. Required the 
distance of the fulcrum C from A. 

Let AC=x; then BC = 36 — x. 

By Art. 133, P.AC=W.BC, 
or 3.x= 15(36-z). 

Hence a;=30 inches =AC. 

Ex. 2. On one arm {=p^ of a straight lever of the first 
kind, without weight, a body counterpoises a weight (=a lbs.), 
on the other (=p 2 ) a weight (=b lbs.). Required the weight 
of the body. 

A straight lever of the first kind, with unequal arms, and hav- 
ing the fulcrum at its center of gravity, is called a false balance. 

Let x= the unknown weight. 

By Art. 133, x.p l =a.p 2 , 

and x.p 2 =b.p 1 . 

By multiplying the equations member by member, 

X ^B \f , 2 =a b-P 'iP ' 2 > 
x 2 =ab, 

x = Vab. 
Ex. 3. On a straight lever, without weight, are suspended 
five bodies, P,=4 lbs., P 2 = 10 lbs., P 3 = 2 lbs., P 4 = 3 lbs., P 5 
= 7 lbs., at the points A, B, C, D, and E, such that AB=4 feet, 
BC=2 feet, CD =6 feet, and DE = 8 feet. Required the posi- 
tion of the fulcrum F, about which they will balance. 

A B C B E 

OPi © 0^3 E OP4 P 5 (J 



84 STATICS. 

Let AF=£. Then, by Art. 136, Cor., we have 

Pj= 4 and/> 1 =AF=.T. •'•Pi?i— 4 - x - 

P 2 = 10 " j9 2 =BF=AF-AB=;c-4. ¥ 2 p 2 = lO(x- 4). 
p 3 = 2 " ^ 3 r=CF=AF-AC=a;-6. Ps/ 3 = 2(x- 6). 
p 4 = 3 " ^ 4 =DF=AF-AD=a;-12. P 4 j» 4 = 3(x-12). 
P 5 = 7 « ^ 5 = EF=AF-AE=:c-20. P 5j p 5 = 7(a;-20). 
And 2.Pp=4x + 10(^-4)+2(o;-6)+3(a;-12)+7(^-20)=0 
=26^—228. 
Hence a;=AF=8if feet. 

Otherwise: Since the weights are parallel forces, their re- 
sultant R is equal to their sum. The whole system being in 
equilibrium, the resultant must pass through the fulcrum, and 
the moment of the resultant must be equal (Art. 45) to the sum 
of the moments of the components. Taking A for the origin 
of moments, we have 

R.ar=P 1 .0+P 2 .AB+P 3 .AC+P 4 .AD+P 5 .AE, 

or 26.z=4X0 + 1X4 + 2X6 +3X12+7X20=228. 
.-. x=AF=8}f feet. 

Ex. 4. A uniform lever AB of the first kind, 12 feet long, 
whose weight w=Q lbs., has a weight W=100 lbs. suspended 
from the shorter arm CB=2 feet. Required the power P 
which must be applied vertically at A, to equilibrate W. 

The weight of the lever has the effect of a weight equal to 
itself, applied at its center of gravity, which, since the lever is 
uniform in size and density, is at its middle point. The dis- 

■ r r, ■ AC-BC 
tance of this point from C is , and its moment about 

tit 

r ~ AC-BC 
C will be w. . 

AC— -RC 
Hence P.AC+w- —-— =W.BC, 

W.BC-| ( AC-BC) 100X2 _ 3X8 
or P= w = ro = 17.6 lbs. 

Ex. 5. The arms of a bent lever are <z=3 feet and b=5 feet, 
and inclined to each other at an angle 0=150°. To the arm 



THE LEVER. 



C3 



a a power P=7 lbs. is applied, and to the arm b a weight W 
= 6 lbs. Required the inclination of each arm to the horizon 
when there is an equilibrium. 



Let a be the inclination of 
the arm a, and (3 the inclina- 
tion of the arm b to the ho- 
rizon. 



ThenP.MC-W.NC, 
or V.a. cos. a=W.5. cos. (3=W.b. cos. (180°- (a+0)), 

= -W.b. cos. (a+6) 

= — W.b. cos. a cos. 0+W.&. sin. a. sin. 0. 
V.a+W.b. cos. 21+30. cos. 1 50° 




■wQ 



Tan. a- 



W.b. sin. S 
21-30 cos. 30° 



30. sin. 150° 
21-30X^3 



4,98 
~15~ 



30. sin. 30 30 X^ 

Log. Tan. «=9.5211381 n , 

fl =-18 .22', 
and ^=18O°-(a+0) = 18O°-l-18 o .22 , -15O o =48 o .22 / . 

Hence the arm AC is inclined at an angle of 18°. 22' above 
the horizon, and BC at an angle of 48°. 22' below the horizon. 

Ex. 6. The whole length of the beam of a false balance 
(Ex. 2) is 3 feet 9 inches. A body placed in one scale coun- 
terpoises a weight of 9 lbs., and in the other a weight of 4 lbs. 
Required the true weight W of the body, and the lengths a and 
b of the shorter and longer arms. 

Ans. W=6 lbs., a=l ft. 6 in., b=2 ft. 3 in. 

Ex. 1. A false balance has one of its arms exceeding the 

other by — th part of the shorter arm. Supposing a shop- 
keeper, in using it, puts the weight as often in one scale as the 
other, does he gain or lose, and how much per cent. ? 

50 

Ans. Loses — t- — per cent. 
m +m 



86 



STATICS. 



Ex. 8. The arms of a bent lever are 3 feet and 5 feet, and 
inclined to each other at an angle of 150°, and to the shorter 
arm is suspended a weight of 7 lbs. Find what the other 
weight must be in order, 1st, that the shorter, and 2d, that the 
longer arm may rest in a horizontal position. 

Ans. 1st, 4.85 lbs, 
2d, 3.64 lbs. 



§11. 



WHEEL AND AXLE. 




137. The Wheel and Axle consists of a 
cylinder or axle, perpendicular to which is 
firmly fixed a circle or wheel, whose center 
is in the axis of the cylinder. The whole is 
supposed to be perfectly rigid, and movable 
only round the axis of the cylinder. 

In the ordinary applications of this ma- 
chine, the power is applied tangentially to 

the surface of the wheel, and the weight, in the same manner, 

to the surface of the axle. 

138. Prop. Required the condition of equilibrium of the wheel 
and axle when two forces are applied tangentially to the circum- 
ference of the axle and of the wheel. 

Let ADE be a section of the 
wheel perpendicular to the axis, 
and BCA a horizontal line 
through the center of the axis, 
terminated at A by the circum- 
ference of the wheel, and at B 
by the circumference of the 
axle. The power P acts in this 
vertical plane at A, and the 
weight W acts in a plane par- 
allel to it. Since the axle and 
wheel are firmly connected, the 
action of the weight may be 
transferred to the plane ADE 




WHEELANDAXLE. 87 

(Art. 57), and supposed to act at B. Then, since the sum of 
the moments must be zero for equilibrium, 

P.AC-W.BC=0. 

W AC radius of wheel 
P ~BC~ radius of axle ' 

or, the Power : Weight = radius of axle : radius of wheel. 

139. Cor. 1. The same relation will exist in all positions of 
the wheel, so that this machine may be called a perpetual lever. 

140. Cor. 2. When the power and weight act vertically on 
opposite sides of the axis, the pressure on the rests or Ys 
equals the sum of the two ; when vertically on the same side, 
it equals their difference ; when in any other directions, it 
equals the diagonal of a parallelogram, whose sides represent 
the power and weight in magnitude and direction. 

If ropes are used to transmit the action of the power and 
weight, we must suppose the forces applied to the axes of the 
ropes. Hence, if r and R represent the radii of the axle and 
wheel respectively, and t and T represent half the thickness 
of the ropes, 

W.(r+0=P.(R+T), 
or P : W=r+t : R + T. 

141. Prop. Required the condition of equilibrium of any num- 
ber of forces, acting in any direction in pla,nes perpendicular to 
the axis. 

Let the actions of all the forces be transferred (Art. 57) to 
the same plane ; then, since the intersection of the axis with 
this plane is a fixed point, the condition of equilibrium is given 
by Art. 74 ; or, the algebraic sum of the moments of all the forces 
about the axis must be zero. 

142. Def. Toothed or cogged wheels are those on the circum- 
ference of which are projections, called teeth or cogs. When 
two wheels have their cogs of such form and distance that 
these of one will work between those of the other, the motion 
of one wheel will be communicated to the other by the pressure 
of the cogs. 



88 



STATICS. 



When the teeth are on the sides of the wheel instead of the 
circumference, they are called crown wheels. 

In the preceding instance, the axes of the wheels and pin- 
ions are parallel or perpendicular to each other. When the 
axes of two wheels make an acute angle, the w T heels take the 
form of frusta of cones, and are called beveled wheels. 

Axles on which teeth are formed are called pinions, and the 
teeth leaves. 

143. Prop. Required the condition of equilibrium when the 
action of the power is transmitted to the weight by means of 
cogged wheels. 

Let S be the mutual 
pressure of one cog upon 
the other. This pressure 
takes place in the direction 
of the line Sm, m'S normal 
to the cogs at the point of 
contact, Cm, Cm' being per- 
pendiculars from the centers 
C and C, of the wheels, on 
that line. 

Taking the moments about C, when the power and weight 
are in equilibrium, we have 

P.C / A=$.C'm', 
and about C, W.CB=S.Cm. 

Dividing the latter by the former, we have 
W_C^A Cm 
P "~ CB "OW 
Now, if the radii of the axles are equal, or C'A=CB, we 
shall have 

WCm 

P ~C'm > . 

which gives the effect of the action of the cogged wheels alone; 

or, since the triangles C'm'o and Cmo are similar, 

Cm _Co_W 

(ym~~C'o~ P ' 




WHEEL AND AXLE. 



89 



If the direction of the line Sm' mS changes as the action 
passes from one cog to the succeeding, the point o will also 
change its position, and the relation of W to P become va- 
riable. 

But when the cogs are of such form that the normal Sm m'B 
at their point of contact shall always be tangent to both circles, 
the lines Cm and Cm' will become radii, and their ratio con- 
stant, and the point o a fixed point, in which case 

WCoCm _R 

T~C~o~Cm 7 ~R" 

R and R' being the radii of the circles C and C respectively. 

144. Cor. When the cogs are equal in breadth, the number 
of cogs on C will be to the number of cogs on C as the cir- 
cumference of C to the circumference of C, or as the radius 
of C to the radius of C. 

W number of cogs on the wheel of W 
P number of cogs on the wheel of P * 

Scholium. When the working sides of the cogs have the 
form of the involute of the circle on which they are raised, the 
pressure of one cog on another will always be in the direction 
of the common tangent to the two wheels. 

Thus, let IHF, KE6 be the two wheels. The acting face 
GCH of the cog a being formed by 
the extremity H of the flexible line 
FaH as it unwinds from the circum- 
ference, and the acting face of b by 
the unwinding of the thread GE, the 
line FCE will always be normal to 
the faces of the cogs a and b at their 
point of contact. The circles describ- 
ed with the radii AD and BD are called the pitch lines of the 
wheels, and will roll uniformly upon each other. 

145. Prop. Required the condition of equilibrium when the 
action of the power is transmitted to the weight by a system of 
cogged wheels and pinions. 




90 



STATICS. 




Let R, Rj, R 2 , &c, be the radii of 



the successive wheels ; 



r 2 ,&c.,the 



radii of the corresponding pinions ; P, 
Pu P 2 , &c, the powers applied to the 
circumferences of the successive wheels. 
Taking the moments about the center 
of each wheel, we have 

RR=P 1 r, P 1 R 1 =P a r 1 ,P a R a =P 3 r 2 ,&c; 

since the power applied to the circumference of the second 
wheel is equal to the reaction on the first pinion. 

Multiplying these equations member by member, and re- 
ducing, 

P n _ R.R 1 .R 2 ,&c. 

P r.rj.rj, &c'i 

or, the power is to the weight as the product of the radii of the 

pinions to the product of the radii of the wheels ; 

or, as the product of the numbers expressing the leaves of each 

pinion to the product of the numbers expressing the cogs in each 

wheel. 



EXAMPLES. 

Ex. 1. If a power of 10 lbs. balance a weight of 240 lbs. on 
a wheel whose diameter is 4 yards, required the radius of the 
axle, the thickness of the ropes being neglected. 

Ans. r=S inches. 

Ex. 2. The radius of the wheel being 2 feet, and of the 
axle 5 inches, and the thickness of each rope being J inch, 
find what power will balance a weight of 130 lbs. 

Ans. P=28§ lbs. 

Ex. 3. The radius of the wheel being 3 feet and of the axle 
3 inches, find what weight will be supported by a power of 
120 lbs., the thickness of the rope coiled around the axle being 
one inch. 

Ans. W=1234f. 

Ex. 4. There are two wheels on the same axle ; the diam- 
eter of one is 5 feet, that of the other 4 feet, and the diameter 



THE CORD. 91 

of the axle is 20 inches. What weight on the axle would be 
supported by forces equal to 48 lbs. and 50 lbs. on the larger 

Ans. W=264lbs. 



and smaller wheels respectively 



§ III. THE CORD. 

146. The cord or rope is employed as a means of communi- 
cation of force. It is regarded as perfectly flexible and with- 
out weight, and transmits the action of a force applied at one 
extremity to any other point in it, unchanged in magnitude, so 
long as it is straight, or only passes over smooth obstacles 
without friction. 

The force thus transmitted is called the tension. 

Since the tension is the same throughout, from one extremity 
to the other, when employed alone, it affords no mechanical 
advantage ; but when passed over or attached to certain fixed 
points, the resistance of these points may be employed advan- 
tageously. 

147. Prop. Required the condition of equilibrium of a cord 
acted upon by three forces. 

Let the forces P 15 P 2 , 
P 3 be applied at the ex- 
tremities A and B of the 
cord ACB, and to a knot 
at C. Draw any line 
CD in the direction of 
the force P 2 , and DN, 
DM parallel" to CA and 
CB respectively. In 
case of equilibrium (Art. 
28), 

CM:CD:CN=P i: P 2 : 




P 3 = sin. DCN : sin. MCN : sin. MCD, 



= sm. a 1 



sin. (a+a') : sin. a ; 
or, the forces are each as the sine of the angle contained between 
the directions of the other tivo. 



92 



STATICS. 



148. Cor. 1. If the cord be fixed at A and B, the reactions 
of the points A and B take the place of the forces Pj and P 3 , 
and are equal to the tensions of the two parts of the cord re- 
spectively. 

149. Cor. 2. If the force P 2 be applied to a running knot or 
ring, the points A and B being fixed, the condition of equilib- 
rium requires that the direction of P 2 should bisect the angle 
ACB. 

For the point C in its motion would describe an ellipse, A 
and B being the foci, and the force P 2 could not be in equilib- 
rium except when normal to the curve, in which case it bisects 
the angle ACB. Hence, sin. a— sin. a', and P a =P 3 ; and since 

Pj : P 2 = sin. a' : sin. (a+a') = sin. a : sin. 2a=l : 2 cos. a. 
.-. P 2 =2P 1 cos. a. 

Otherwise : since the tension of the cord is the same through- 
out, when the cord passes over an obstacle without friction, 
P i= P 3 ; a =a' and P 2 =2P 1 cos. a (Art. 19). 

150. Prop. Required the conditions of equilibrium when any 
number of forces in the same plane are applied at different points 
of the cord. 

Let ABCDE be a cord to which are applied the forces P t , 

P 2 , P 3 , &c, at the 
points A, B, C, &c, in 
the directions AP T , 
BP 2 ,CP 3 , &c. 

The force P, at A 
may be considered as 
acting at B in its di- 
rection BA ; and since 
B, when in equilibrium, 
must be so from the 
action of P l9 P 2 , and 
the tension of BC, the 
resultant of P, and P 2 must be in the direction of CB, and 
may be considered as acting at C. Suppose it thus applied, 
and let it be resolved into two, acting in the directions Cn and 
Cm parallel to the original components, and equal to them. 




THE CORD. 



93 



We have thus transferred the forces Pj and P 2 to act at C 
parallel to their original directions. In the same manner, the 
resultant of P 15 P 2 , and P 3 acting at C must be in the direction 
of DC, and may be applied at the point D without disturbing 
the equilibrium, and then replaced by P,, P 2 , and P 3 parallel to 
their original directions, and so on, for any number. Hence, 
if all the forces be supposed to act at one point parallel to their 
original directions, they will be in equilibrium. The conditions 
of equilibrium, therefore, are the same as for any number of 
concurring forces (Art. 70), or, the sum of all the forces re- 
solved in any two rectangular directions must be zero. 

The form which the cord takes under the influence of the 
several forces is called a, funicular polygon. 

151. Prop. Required the relations of the forces which, acting 
on a cord in one plane, keep it in equilibrium. 

Produce P 2 B to N, 
P 3 C to M, &c, and let 
the 

Z-ABN=«, z_NBC=a', 

Z-BCM=ft z_MCD=0\ 

&c., &c. 

Let t\, t z , t 3 , &c, be 
the tensions of the sev- 
eral successive portions 
of the cord. 




Then (Art. 147) t x : P 2 : 

also, t 2 : P 3 : 

&c, 



t 2 = sin. a 



sin. (a-\-a') : sin. a; (a) 
t 3 = sin. (3' : sin. ((3+(3') : sin. j3, (a 1 ) 
&c. 



From (a) we obtain t^=V 
" (a 1 ) 



sin. a' 



2 sin. (a-\-a') 

8JD.0 

2 rs, sin.(j3+i3 / ) 
&c, 



, and t 2 =~P, 



sin. a 



■jr, and£ 3 =P 



sin. (a+a 1 )' 

sin. (3 

sin.^/*')' 

&c. 



Equating the values of t 2 , t^ &c, we have 
p sin. a _ sin. (3' 

2 *sin. (a+a')~ 3 'sin. (^+^)' 



© 



94 



STATIC 



sin. (3 



-=P-.- 



sin. y' 



m 



3 'sin. (j3+j3') ^ 4 'sin.(y+y')' 
&c, dec, 

for the relations of the forces. 

152. Cor. 1. If the cord be fixed at the points A and E, and 
the forces P 2 , P 3 , P 4 , &c, be parallel, we have 
sin. a'= sin. |3, sin ,f3'= sin. y, &c. 
Multiplying these equations by (6), (&') member by member 
in their order, we obtain 

sin. a sin. a' _ sin. /3 sin. |3'_ sin. y sin. y' 
2 *sin. (a+a')~ 3 ~sin. {P+P') = 4 sin. (y+y')' 
sin. a sin. a' sin. a sin. a' 



But 



sin. (a+a') sin. <z cos. a' + sin. a' cos. a 
1 1 



cos. a cos. a' cot.a+cot. a'' 
sin. a sin. a' 



cot. a+ cot. a' cot. ,3+ cot. /3' cot. y+cot. y' 



&c. 



153. Cor. 2. If the cord be 
fixed at A and E, and the forces 
P 2 , P 3 , P 4 , &c, be weights, the 
horizontal tension of each por- 
tion of the cord is the same. 

For, by resolving the tension 
of each part horizontally, we 
have 

the horizontal tension of AB=^. sin. ABN=^ sin. a, 

BC=t 2 sin.BCM=£ 2 sin.fr&c. 
Substituting for t t9 t 2 , &c, their values found above, 




horizontal tension of AB=P 



BC=P 



sin. a sin. a: 



sin. (a+a') cot. a + cot. a' 
sin. /3 sin. /3' P 



sin.(0+j3') cot. 0+ cot. 0" 
which, by Cor. 1, are all equal. 



T H E C O R D. 95 

154. Cor. 3. Since the reactions of the points A and E equi- 
librate the resultant of all the weights, the lines AB, ED pro- 
duced, must meet in some point of the vertical through the 
center of gravity of the system. For three forces, if in equi- 
librium, must meet in a point. 

155. A heavy cord may be considered a funicular polygon 
loaded with an infinite number of small weights, and since the 
number of weights is infinite, the polygon will also have an 
infinite number of sides, or will become a curve. 

The curve which a heavy cord or chain of indefinitely small 
links will assume, when suspended from two fixed points not in 
the same vertical line, is called the catenary. 

EXAMPLES. 

Ex. 1. Two equal weights balance, by a cord, over any 
number of fixed points without friction. Required the press- 
ure on each. 

Since the tension of the cord P a ABCDEP r 
throughout, each point is acted upon 
by three forces, viz., two equal ten- 
sions on each side of it and the reac- 
tion of the point, which last must be 
equal to the resultant of the other two. 
Hence, calling the angles at A, B, C, 
&c, a, b, c, &c, by Art. 15, and Cor., 
Art. 19, 

the pressure on A=2P T cos. \a, 
B = 2P 1 cos.|&, 
&c, &c. 

Ex. 2. A cord of given length passes over two fixed points 
A and B without friction, and one extremity, to which a given 
weight P is attached, passes through a small ring at the other 
extremity C. It is required to find the tension of the cord when 
in equilibrium, and the length of the part CP below the ring. 

Since the cord passes freely over the points A and B, and 
through the ring C, it is of the same tension throughout, and 
equal to the weight P. Hence the point C is kept at rest by 




96 



STATICS. 




B three equal forces, and, by Art. 18, must make 
angles of 120° with each other. Draw the 
horizontal line AD; and, since ACD=120°, 
ADC and CAD are each equal to 30°. The 
position of A and B being given, the angle 
BAD must be known. Hence, in the triangle ACB we have 
the side AB and all the angles from which AC and BC may- 
be determined. Then the whole length of the cord, diminished 
by the perimeter of the triangle, will be the distance of the 
weight from the ring. 



§ IV. THE PULLEY. 

156. The pulley is a small grooved wheel movable about 
an axis, and fixed in a block. The cord passes over the cir- 
cumference of the wheel in the groove. 

The use of the pulley is to prevent the effects of friction and 
rigidity of the cord. The first of these it diminishes by trans- 
ferring the friction from the cord and circumference of the 
wheel to the axle and its supports, which may be highly pol- 
ished or lubricated. The effects of rigidity are diminished by 
turning the cord in a circular arc instead of a sharp angle. 

The pulley is called fixed or movable, according as the block 
is fixed or movable. 



157. The fixed pulley serves merely to 
change the direction of the forces trans- 
mitted by the cord, since, neglecting the 
friction of the pulley, the tension of the 
cord is the same in every part of it. Hence 
the power equals the weight, and the press- 
ure on the axis of the pulley equals their 
sum. 




THE PULLEY. 



97 



158. Prop. Required the relation of the power to the weight in 
the single movable pulley. 

The tension t of the cord, being the same p 
throughout, is equal to the power P ; also 
to the pressure on the hook Q,. 

The resultant of the two tensions in the 
directions CP and DQ, being equal and 
opposite to the weight W, must be vert- 
ical. Let a be the angle made by the 
cords with this vertical, and, resolving the 
tensions vertically, we have 

2t cos. a=2Y cos. a=W. 
W 




2 cos. a 
the same as obtained in Art. 149. 

Cor. 1. If the weight w of the pulley be taken into account, 

E= w±». 

2 cos. a' 
Cor. 2. If the cords are parallel, a—Q and 

„ W+w 



Cor. 3. If a=90°, 2«=180°, or the cord becomes straight 
and horizontal. In this case 

-o W+w 
P=-— =«, 

or the power must be infinite. In other words, no power can 
reduce the cord to a horizontal straight line while the weight 
is finite. 

159. Of the various combinations of pulleys there are three, 
which we shall distinguish by the first, second, and third sys- 
tems of pulleys. 

160. Prop. Required the relation of the power to the weight in 
the first system of pulleys. 

G 



98 



STATICS. 




The annexed figure represents this system. 
Neglecting friction, the tension t of the cord 
is the same throughout, and equal to P. The 
weight W and the weight of the lower block 
w are sustained by the tensions of the several 
cords at the lower block. Hence, if n be the 
number of cords at this block, 
nt= n Y=W+w. 

n 
If the weight of the lower block be neg- 
lected, 

n 
Of this system of pulleys there are various 
modifications. The annexed form is the one 
in most common use. 



161. Prop. Required the relation of the power to the weight in 
the second system of pulleys. 

The annexed figure represents this system with three mov- 
able pulleys, each pulley having its own rope. 
Designate by «,, a 2 , a 3 , &c, the pulleys re- 
spectively in their order from the weight W, 
by io,, w 2 , w 3 , &c, their weights, and by t lt 
t 2 , t 3 , &c, the tensions of their respective 
cords. 

Then, for the equilibrium of a,, we have 

2J 1 =W+w 1 , or *,== -— . 

W+w 1 -\-2w 2 
For the equilibrium of a 2 , 2t 2 —t l +w 2 = ;- -, 

W+w l J r 2w 2 




or 



* ^ '■.'. , W+w 1 +2w 9 +2*w. 
For the equilibrium of a 3 , 2t 3 =t 2 -\-w 3 = — — — 



or 



THE PULLEY. 

W+w l +2w 2 +2*w, 



For the equilibrium of a n , 

W+w 1 +2w 2 +2' 2 w :i +2'Wi+ . . . 2 r 



P=t H =- 



l w n 



Cor. 1. If the weight of each pulley is the same, and equal w x , 

v= W+w, (l+2+2 2 +2 3 + .... 2- 1 ) 
2 n 



W w 

— i 


(2"- 


"I) 


- 2 .+ 


2" 




w 


■(- 


1 

~2" 



Cor. 2. If the weights of the pulleys be neglected, 

W 
P= * or YV=2 W .P. 

162. Prop. Required the ?~elation of the power to the weight 
in the third system of pulleys. 

In this system each cord is attached to the 
weight, and the number of movable pulleys is one 
less than the number of cords. Designating the 
pulleys in their order from the weight by a,, a 2 , 
a 3 , &c, their weights respectively by w 19 w 2 , 
w 3 ,&lc, and the tensions of the successive cords 

by*,, 



2 , £3, »-» 




&c, we have 

t 2 =2t 1 +w 1 =2¥+w 1 , 

t 3 =2t 2 +w 2 =2*~P+2w l +w 2f 

t 4 =2t 3 +w 3 =2 5 ¥+2 2 w 1 +2w 2 +w 5 . 



And if there be n cords, 

t n =2"- 1 V+2"-hv 1 +2 n -*w 2 + &c 2w n _,+w n _ 1 . 

ButW=« 1 +^+^ 3 + ,&c t n 

=P(l+2+2 2 +2 3 +,&c....2"- 1 )+i^ 1 (l+2+2 2 +,&c 
. . . 2- 2 )+w; 2 (l+2+2 2 +,&c 2"- 3 )+,&c. 



100 



STATICS. 



= P(2 B -l)+tfl x (2^ 1 -l)H-!i> a (2^-l)+,&c 

10^(2-1), 

If the (n— 1) pulleys are of the same weight w l , 

W=P(2"-l)+w; 1 (2"- 1 +2 n - 3 +2' J - 3 +, &c 2 a +2 + l-rc) 

=V(2 n -l)+w 1 (2"-l)-w 1 n 
= (P+w 1 )(2 n -l)-w 1 n. 
If the weights of the pulleys be neglected, 
W-P(2 n -1). 



EXAMPLES. 

Ex. 1. At what angle must the cords of a single movable 
pulley be inclined in order that P may equal W ? 

Ex. 2. In the first system of pulleys, if there be 10 cords at 
the lower block, what power will support a weight of 1000 lbs. ? 

Ex. 3. In the second system of pulleys, if 1 lb. support a 
weight of 128 lbs., required the number of pulleys supposed 
without weight. 

Ex. 4. In the third system of 6 pulleys, each weighing 1 lb., 
find what weight will be supported by a power of 12 lbs. 
j 



Ex. 5. Find the ratio of the power to the 
weight in the annexed modification of the second 
system of pulleys. 




§ V. THE INCLINED PLANE. 

163. The Inclined plane, as a mechanical power, is supposed 
perfectly hard and smooth, unless friction be considered. It 
assists in sustaining a heavy body by its reaction. This reac- 
tion, however, being normal to the plane, can not entirely 
counteract the weight of the body, which acts vertically down- 



THE INCLINED PLANE. 



101 



ward. Some other force must therefore be made to act upon 
the body, in order that it may be sustained. 

164. Prop. Required the conditions of equilibrium of a body 
sustained by any force on an inclined plane. 

Let AB be a section of an inclined plane, of which AB is 
the length, BC the 
height, and AC the 
base. Let i be the in- 
clination of the plane to 
the horizon, e the angle 
made by the direction 
of the power P with the 
plane AB, W= the 
weight of the body «, 
and R= the reaction of 
the plane. The body is kept at rest by the action of P, W, 
and R. Resolving the forces parallel and perpendicular to 
the plane, we have 

Pcos. e — W sin. i=0, (a) 

R + P sin. e- W cos. i=0. (6) 

From (a) we obtain 

W COS. £ 




sin. i 



From (6), R=W cos. i— P sin. e=W cos. i— 



W sin. t sin. £ 



cos. e 



or 



^ xos.(i+e) 

COS. £ 
W COS. £ 



(d) 



R cos. (i+e) 9 
the same relations as obtained in Ex. 24, Art. 129. 

165. Cor. 1. If the force P act parallel to the plane, £=0, 
and (c) becomes 

W_ 1 AB 
P~sin.z~BC' 
or, the power is to the weight as the height of the plane to its 
length. 



102 



STATICS. 



t? /a W 1 AB 

I rom (d) we get -^= ;=t^, 

v ' & R cos. % AC 

or, the reaction of the plane is to the weight as the base to the 
length. 

166. Cor. 2. If the power act parallel to the base of the plane, 
e= — i, and (c) becomes 

W_cos. z'_AC 

"P~"sTnT7~BC' 
or, the power is to the weight as the height to the base. 
_, ... W cos.; AC 

From (d), TT=-r = AB' 

or, the reaction of the plane is to the weight as the length to the 
base. 

167. Prop. Required the conditions of equilibrium of two 
bodies resting on two inclined planes having a common summit, 
the bodies being connected by a cord passing over a pulley at 
the summit. 




Let W and W, be the 

weights of the bodies, and 
i, i t the inclinations of the 
planes. 

If t be the tension of the 
cord, we have for equilib- 
rium on the plane AB 
c (Art. 165), 



t=W. sin. i, 
t=W , sin. z'j. 
. Wsin. i=W, sin. i x , 

V AB VVl BC 
WAB 
W^BC 

or, the weights are proportional to the lengths of the planes on 
which they rest respectively. 



on the plane BC 



or 



or 



THE WEDGE. 



103 



EXAMPLES. 

Ex. 1. What force acting parallel to the base of the plane 
is necessary to support a weight of 50 lbs. on a plane inclined 
at an angle of 15° to the horizon? 

Ex. 2. If the weight, power, and reaction of the plane are 
respectively as the numbers 25, 16, and 10, find the inclination 
of the plane, and the inclination of the power's direction to 
the plane. 



§ VI. THE WEDGE. 

168. The wedge is a triangular prism whose perpendicular 
section is an isosceles triangle. The dihedral angle formed 
by the two equal rectangular faces, is called the angle of the 
wedge. The other rectangular face is called the back. It is 
used to separate the parts of bodies, by introducing the angle 
of the wedge between them by a power applied perpendicu- 
larly to the back. The equal rectangular faces are regarded 
as perfectly smooth, in which case the only effective part of 
the resistance must be perpendicular to these faces. 

169. Prop. To determine the conditions of equilibrium in the 
wedge. 

Let ABC be a section of the 
wedge perpendicular to the angle 
or edge A. Draw AD bisecting 
the angle, and let BAD = CAD=«. 
Let 2P be the power applied to the 
back BC of the wedge, which must 
be in equilibrium with the pressures 
R on the two faces AB and AC. 
If an equilibrium exist, the forces 
2P, R, and R must meet at some 
point in AD (Art. 74). 

Resolving the forces vertically, 
or in the direction AD, we have 




104 



STATICS. 



or 



or 



R -sin.«- BA - BA/ 



2R.sin.BAD-2P=0, 
P=R. sin. «, 
BD BD,Z \ the back of the wedge 
face of the wedge ' 

where 1= the length of the edge or breadth of the face. 

170. In the foregoing investigation of the theory of equilib- 
rium in the wedge, we have omitted the consideration of the 
friction, and have supposed the power to be a pressure ; 
whereas, in practice, the wedge is kept at rest by friction alone, 
and the power arises from percussion. The following prob- 
lem will serve to elucidate the theoretical view here taken of 
the wedge. 

171. Prob. A heavy beam is attached, by a hinge at one 
end, to a smooth, horizontal plane, while the other rests on the 
smooth face of a semi-wedge. Required the horizontal force 
necessary to keep the wedge from moving. 

Let DE be the beam 
and BAC the wedge, 
i.BAC = a, z.ADE = ft 

/= the length of the 
beam, g the center of 
gravity, and Dg=a. 

The wedge is kept in equilibrium by the pressure of the 
beam upon it at E, and the horizontal force P acting upon it 
at some point H. The beam is kept in equilibrium by its 
weight w acting at g and the reaction R of the face of the 
wedge at E. 

Taking the moments about D for the equilibrium of the beam, 
we avoid expressions involving the unknown thrust R', and 
have 

w.Dg. cos. ADE-R.DE. sin. DER=0, 
or w.a. cos. j3— R.Z. cos. (a—(3) = 0. 




R: 



a 



cos. j3 



/'cos. (a—0)' 
But the principles of the wedge give 
P=R. sin. a. 



THE SCREW. 



105 



_> a sin. a cos. (3 

.'. P=w. 7 . — . 

/ cos. (a— (3) 

By an examination of this value of P, it will be seen that the 

power necessary to keep the wedge from moving will diminish 

as the wedge advances beneath the beam. 



§ VII. 



THE SCREW. 




172. If we divide the rectangle ABCD into equal parts by 
the lines mn, m'n', &c, parallel 
to AB, and draw the parallel di- 
agonals of the rectangles thus 
formed, and if we suppose the 
whole rectangle to be wrapped 
round the surface of a cylinder, 
the perimeter of whose base is 
equal to AB and altitude to BC, 
the diagonals of the rectangles 
will trace on the surface of the cylinder a continuous curve, 
which is called the helix. 

If a projecting thread or rib be attached to the cylinder upon 
this curve, we have the screw, sometimes called the external 
screw. Similarly, if we take a hollow cylinder of exactly the 
same radius as the solid one, and generate a groove in the 
same curve, we have the internal screw or nut. 

The screw works in the nut, either of which may be fixed 
and the other movable. 

173. Prop. To determine the conditions of equilibrium in the 
screw. 

From the construction of the 
screw, it appears that the thread 
of it is an inclined plane, of 
which the base is the circum- 
ference of the cylinder and the 
height the distance between the 
threads. The force is generally 
applied perpendicularly to the 
end of a lever inserted into the 




106 STATICS. 

cylinder, and in the plane perpendicular to the axis of the 
cylinder. The power P thus applied, in turning the screw 
round, produces a pressure on the threads of the screw in the 
direction of the axis of the cylinder. In case of equilibrium, 
let the counterpoise of this pressure be W. Let ABC be the 
inclined plane formed by unwrapping one revolution of a 

thread, and let w=-th part of W, be supported at «, q being 

the same part of Q, the force applied at the circumference of 
the cylinder. Put FD = a, ED the radius of the cylinder =r, 
and angle BAC=a. Then, Art. 133, 

ED r 2irr 
By Art. 166, q—w. 



or 



AC 

Q, W distance of two threads 



n n ' circumference of the cylinder' 
and the same holds for each of the other portions of W at the 
other points of the plane. Therefore, we have 
distance of two threads 
' circumference of the cylinder* 
2?ra_ circumference described by P 
y v )f H— '2irr~ ' circumference of cylinder 
W circumference described by P 
P _ distance of two threads ' 
or, the power is to the weight as the distance of two threads 
is to the circumference described by the power in one revolu- 
tion of the screw. 

It will be seen that the ratio of P to W is independent of 
the radius of the cylinder. 

examples. . 

Ex. 1. What force must be exerted to sustain a ton weight 
on a screw, the thread of which makes 150 turns in the height 
of 1 foot, the length of the arm being 6 feet ? 

Ex. 2. Find the weight that can be sustained by a power 



BALANCES, ETC. 



107 



of 1 lb. acting at the distance of 3 yards from the axis of the 
screw, the distance between the threads being 1 inch. 

Ex. 3. What must be the length of a lever at whose ex- 
tremity a force of 1 lb. will support a weight of 1000 lbs. on a 
screw, whose threads are f inch apart? 



§ VIII. BALANCES AND COMBINATIONS OF THE MECHANICAL 
POWERS. 

174. The common balance, in its best form, is a bent lever, 
in which the weight of the lever must be taken into consider- 
ation. 

In the annexed figure the points A and B, from which the 
scale-pans and weights 
are suspended, are call- 
ed the points of suspen- 
sion ; C is the fulcrum, 
being the lower edge 
of a prismatic rod of 
steel projecting on each 
side of the beam; when 
the balance is in use, 
these edges on each 
side of the beam, as at 
C, rest on hard surfa- 
ces, so that the beam turns freely about C as a fulcrum. 

175. The requisites of a good balance are, 

1°. That the beam rest in a horizontal position when loaded 
with equal weights. 

2°. That the balance possess great sensibility. 
3°. That it possess great stability. 

176. Prop. To determine the conditions that the beam rest in 
a horizontal position when loaded with equal weights. 

Supposing the beam horizontal, in case of equilibrium, if we 
neglect the weight of the beam, the moments of the weights 
must be equal, and, therefore, the arms must be equal {Art. 133). 




108 



T AT IC 



But taking into consideration the weight of the beam, its cen- 
ter of gravity must be in the vertical through C, the center of 
motion (Art. 110), and, in order to this, the beam must be sym- 
metrical on opposite sides of the fulcrum (Art. 97). The line 
AB, joining the points of suspension, is obviously bisected by 
the vertical through C and the center of gravity of the beam, 
and the point of intersection, for reasons which will appear, 
should be below C. 

177. Prop. To determine the conditions that the balance may 
possess great sensibility. 

Let C, A, and B be the fulcrum and points of suspension, as 

-g in the preceding figure, and 

_ , ^*- e *"1 ioin Cg, the centers of mo- 

M Cda ^^e\ AT r. \ c n • 

tion and of gravity. Lg is 

perpendicular to AB and bi- 
sects it, if the beam is con- 
structed in accordance with 
the first requisite. Let M 
C E N be a horizontal line 

W-p r\ * 

T ^ through C, meeting AB in E, 

making with it an angle equal to 6. 

The sensibility is measured by the amount of deflection 6 
of the line AB from a horizontal position by a given small dif- 
ference P— Q of the weights. 

Draw the vertical lines Dd and ga, and put AD=BD=«, 
CD=d, Cg=h, and weight of the beam =w. Now MN is bi- 
sected in d, and Md=a cos. 6, Cd=d. sin. 6, and Ca=h. sin. 6. 

If the system is in equilibrium, the moments about C give 

P.CM-Q.CN-^.Ca=0, 

or V.(Md-Cd)-Q,(Nd+Cd)-w.Ca=0, 

or (P-Q)a. cos. 0.-(P+Q)tf sin. 6-w.h. sin. 0=0, 

or (P-Q)a-{(P+Q).d+u>.A} tan. 6=0. 

, a _ (P-Q)-a 
•'' tan ' e/ -(P+Q)rf+u>.A' 

Hence the angle 6, and, therefore, the sensibility, is increased 
for given values of P and Q, by increasing the lengths of the 




BALANCES, ETC. 109 

arms (a), by diminishing the weight of the beam (w), or by di- 
minishing the distances of the fulcrum from the center of grav- 
ity of the beam (h) and from the line joining the points of sus- 
pension (d). 

178. Prop. To determine the conditions that the balance may 
possess great stability. 

If the balance be loaded with equal weights and disturbed 
from its position of equilibrium, the rapidity with which it re- 
turns to that position is a measure of its stability. But this 
rapidity of return to a horizontal position will depend upon the 
moment which urges it back. 

But this moment is, since P=Q, 

P.CN-P.CM+u>.Ca, 

or ¥.2.Cd+w.Ca, 

or {2?.d+w.h) sin. 6. 

Hence, for given values of P and 6 the stability is greater, 
as d, h, and w are increased. 

179. Cor. Hence, by increasing the stability, we diminish 
the sensibility, but the sensibility may be increased by increas- 
ing the length of the arms, without affecting the stability. 

For commercial purposes, when expedition is required, and 
the material weighed is not of great value, sensibility is sacri- 
ficed to stability ; but for philosophical purposes great sensi- 
bility is required, and stability is of little comparative import- 
ance. 

THE STEELYARD BALANCE. 

180. The steelyard balance, or Roman steelyard, is a lever 
of the first kind, with un- r, 

equal arms. The body M , q 

W to be weighed is hung A l? ' \T? t 1 n >* g £ — 3 B 

at the shorter arm A, and jf p 

a given constant weight 3L 

P is moved along the ^P 

other arm till it balances W ; then the weight of W is known 

from the place of the counterpoise P. 



110 



STATICS. 



181. Prop. To determine the law according to which the 
longer arm of the steelyard must be graduated. 

Let G be the center of gravity of the beam AB and w its 
weight. Put OA=p, 00=^, OG=g, and taking moments 
about the fulcrum O, in case of equilibrium we have 
Wp—wg—'Pp 1 =0. 
wg W» 

Taking Ox, a fourth proportional to P, w, and g, we have 

Hence the distance from x to the counterpoise P varies as 
the weight ; and if the weights be in arithmetical progression, 
the distances xl, x2, xS, &c, will also be in arithmetical pro- 
gression. 



THE BENT LEVER BALANCE. 




182. This balance is represented by 
the annexed figure, where ABC is the 
bent lever turning about a pivot at B. 
A scale (E) hangs from A, and at C an 
index points to some division on the 
graduated arc GCF. 



183. Prop. To determine the principle of graduation in the 
bent lever balance. 

Let g be the center of gravity of the beam at which its 
weight w acts. The weight of the scale (E) and the weight 
(W) of a body placed in it will act vertically through A. Let 
the horizontal line through B meet the vertical lines through 
g and A respectively in D and K. Then the moments about 
B, in equilibrium, give 

w.DB-(E-fW)BK=0. (a) 



BALANCES, ETC. 



Ill 



As greater weights are put into the scale E, the point A ap- 
proaches more nearly the vertical through B from the bent 
form of the beam, or BK diminishes, while BD increases. 

Suppose the point A to be at K when the scale is unloaded, 
and let, in this position of the beam, the angle DBg=6. When 
a weight is put into the scale the point A will descend through 
some angle 0, and the arm B^- will rise through the same angle. 
In this new position the angle DBg will become 6— 0. Let 
Bg=a and BA=b ; then 

DB=a cos. (#—</>), and BK=b cos. 0. 
These substituted in (a), give 

w.a. cos. (6— 0) — ~E.b. cos. 0— W.b. cos. 0=0, 
or w.a. cos. cos. (p-\-w.a sin. sin. 0— E.fr cos. 0— W.b. cos. 0=0, 
or w.a. cos. d-\-w.a sin. 6 tan. — E.6— W.b=0. 



tan. 0= 



W.b E.b — w.acos.O 



i+- 



w.a. sin. v w.a. sin. 6 

Hence, tan. varies as W, and the limb GF must be di- 
vided into arcs whose tangents are in arithmetical progression. 

Practically, the limb may be graduated from the positions 
of the index at C for a succession of weights put into E. This 
instrument possesses great stability. 



ROBERVAL S BALANCE. 



184. This instrument is of greater interest from its paradox- 
ical appearance than from its use as a machine for weighing 
bodies. Its discussion affords an interesting application of the 
doctrine of couples. 

It consists of an upright 
stenvupon a heavy base A, 
with equal cross-beams 
turning about pivots at a 
and b. These cross-beams 
are connected by pivots at 
c, d, e, and/, with two oth- 
er equal pieces in the form 



f 



112 



T A T I C S. 



Q2^ 




of a T. The weights are suspended from the horizontal arms 
of the latter pieces. 

185. Prop. In RobervaPs balance, equal weights balance at all 
distances from the upright stem. 

Let the letters in the 
annexed figure indi- 
cate the same parts as 
in the former. 

Let equal and oppo- 
site forces Pj and P 2 
each equal to P act in 
ec, and, similarly, let 
P 3 and P 4 act in df. 
These forces P,, P 2 , 
P 3 , and P 4 do not dis- 
turb the equilibrium. Now P at B, as in Art. 71, is equivalent 
to Pj at e and the couple P, B^, P 2 , and, similarly, P at C is 
equivalent to P 4 at / and the couple P, Ck, P 3 ; and since P 1 
at e balances P 4 at /, we have only the two couples to dis- 
pose of. 

Now for the couple P, Bh, P 2 , Arts. 58 and 59, we may sub- 
stitute a couple Q,,ec, Q 2 in its own plane and of equal mo- 
ment, in which the forces Qj and Q 2 , acting in the directions 
of the cross-beams cd and ef (which always remain parallel to 
each other as they turn on the pivots a and b), are destroyed by 
the resistance of the pivots a and b. Similarly, the couple P, 
Ck, P 3 may be replaced by the equivalent couple R l ,fd, R 2 , 
in which R x and R 2 are destroyed by the resistance of b and 
a. These new couples, therefore, do not disturb the equilibri- 
um, and the original forces P at B and P at C must be in equi- 
librium. 

If the beams cad and ebf be moved round the pivots into 
any oblique position, the same reasoning would apply, and the 
equilibrium still subsist. 

Cor. Unequal weights can not balance from whatever points 
suspended. 



BALANCES, ETC. 



113 



C" 

\7_ 



B" 



W 



B' 



W / 



C 

J7 



vp 



A' 



w 



186. Prop. To determine the ratio of the power to the weight 
in a combination of levers. 

Let the power P act at A, q 

and the weight W at B'". 
The first three levers are of 
the second kind, and the last 
one of the first kind, the ful- 
crums being at C, C, C", and 
C". Let BA', BA", B"A'" 
be rigid rods connecting the 
levers, and let the action of 
the first lever on the rod BA' 
be W, which becomes the 
power acting on the second, 
W" and W" the weights to the second and third respectively, 
and powers to the third and fourth. 

P CB W C'B' W" C"B" . W" C ;// B ,;/ 



Then ^777=- 



CB" W" 

-, and 



W CA' W ,; C'A" W ,,; C"A'" W C^'A'"' 

Taking the continued product of these equations member 
by member, we have 

P _ CBxC / B'xC // B // xC /// B" / 

W~CAxC'A'xC // A / 'xC /// A /// ; 

or, the ratio of the power to the weight in the combination is 
equal to the product of the ratios of the power to the weight in 
each lever. 

187. Prof. To determine the ratio of the power to the weight 
in the endless screw. 



This machine is a combination of the 
screw and wheel and axle. 

Let P be the power applied to the han- 
dle of the winch, W the pressure of the 
screw on the teeth of the wheel, and W 
the weight suspended from the axle of 
the wheel. Then 



H 




114 STATICS. 

P _ distance between the threads of the screw 

W circumference described by the power ' 

W'_ radius of axle 

W radius of wheel' 

P _distance between two threads of screw 

W circumference described by the power 

radius of axle 
\/ • 

radius of wheel ' 
or, the ratio of the power to the weight in the endless screw is 
equal to the product of the ratios of the power to the weight in 
the screw and in the wheel and axle. 

188. Prop. To determine the ratio of the power to the weight 
in any combination of the mechanical powers. 

Let P= the power for the whole combination, 
W"= " weight " " " 

W = " " to the 1st in the series and power to the 2d, 
W"= " " " 2d " " " 3d, 

&c, &c, &c. 

Let a l = the ratio of the power to the weight in the 1st, 
a 2 = " " " " " 2d, 

a 3 = " " " " " 3d, &c. 

_Z_ W ' - W " - W- l _ 

I hen yy,— a i> \y// — a ^ y\fm~ a z yy n — a »i 

and, taking the product, 

P 
^7=a 1 .a 2 .a 3 a n . 

Hence the ratio of the power to the weight in any combina- 
tion of elementary machines is equal to the product of the ratios 
of the power to the weight in each of the simple machines. 

189. Prop. To determine the ratio of the power to the pressure 
in the combination of levers called the knee. 

This combination of levers is used with advantage where 
very great pressure is required to act through only a very 
small space, as in coining money, in punching holes through 
thick plates of iron, in the printing-press, &c. The lever AB 



BALANCES, ETC. 



115 



^P 




turns about a firmly fixed pivot 
at A, and is connected by anoth- 
er pivot at B to the rod BC, 
whose extremity C produces the 
pressure on the obstacle at E. 

Let the power P act horizon- 
tally at some point F in the le- 
ver AB, ANC be a vertical line 
meeting the direction of P in N, 
and DE a horizontal plane, on 
which, at E, is the substance 
subject to pressure. Let R= 
the reaction of the rod BC in the 
direction of its length, AM, DL perpendiculars upon its direc- 
tion from A and D, and W the vertical resistance of the sub- 
stance at E. 

Taking the moments about A and D in equilibrium, we have 
P.AN=R.AM, and W.DE=R.DL, 
P AMxDE 
or W~ANxDL' 

When BC becomes nearly vertical, DL becomes nearly equal 
to DE, and AN to AF, while AM becomes very small. 

T 1 • ■ P AM , 

In this situation, ™f= ~rw nearly, so that P is a very small 
fraction of W. 



CHAPTER VII. 



APPLICATION OF THE PRINCIPLE OF VIRTUAL VELOCITIES TO THE 
MECHANICAL POWERS. 

190. In Arts. 80 and 81, it is shown that the principle of 
virtual velocities obtains for all cases of equilibrium of a free 
body under the actions of any number of external forces in the 
same plane. 

In the mechanical powers, the parts by which the actions 
of the forces are transmitted being rigid or inextensible, the 
forces may be considered as acting in the same plane, and the 
internal reactions and tensions will not enter the fundamental 
equation 2.P.u=0. Also, the virtual velocities of the support- 
ing parts will in general be zero for the possible displacements 
of the system. 

In some of the mechanical powers, the principle applies to 
all possible displacements, however great, since they must be 
in the direction of the forces. This is true in the wheel and 
axle, toothed wheels, pulleys with parallel cords, the inclined 
plane, the wedge, and screw. In the lever, and pulleys with 

inclined cords, the displace- 
ments must be taken indefinitely 
small. 

191. Prop. The principle of 
virtual velocities obtains in the 
wheel and axle in equilibrium. 

The forces which act on the 
wheel and axle are the power 
P, the weight W, and the reac- 
tion R of the steps which sup- 
port each end, C, of the pivot 
about which it turns, and, in con- 
sequence of the rigidity of the 




PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 117 



system, they may be considered as acting in the same plane. 
Also, the wheel and axle receiving a displacement turning 
about C, the virtual velocity of R equals 0. 

Let A and B be the points at which the cords left the wheel 
and the axle respectively before displacement ; A', B' after- 
ward. Then W ascends through the space WW- arc BB', 
and P descends through PP'=arc A A'. PP' is the virtual ve- 
locity (Art. 78) of P, and positive ; WW' is the virtual veloc- 
ity of W, and negative. Hence (Art. 81), 

p.pp-w.ww'=o, 

or P. arc AA'- W. arc BB'=0, 

or P.AC. z. A'C A- W.BC. Z-BCB'=0. 



P.AC-W.BC: 



W AC 
°' ° r P = BC' 



the condition of equilibrium found in Art. 138. 

192. Prop. The principle of virtual velocities obtains in a 
pair of toothed wheels. 

Let the circles in 
the annexed figure 
represent the pitch- 
lines of the wheels 
(Art. 144),andD 7 ,D 2 
the points which were 
in contact in the line 
CC before displace- 
ment. Since the pitch- 
lines roll on each oth- 
er without slipping, 
arc DD T =arc DD 2 , 
and 

arc D D 
P's displacement =PP'=AC'.z.D 1 C'D=AC'.— ^-, 

W's displacement =WW'=CB.z.D 2 CD=CB. ar 




CD 



By the principle of virtual velocities, 

p.pp'-w.ww=o, 



118 



STATICS. 



^.-.arcDjD T __ .,_. arc D 2 D rt 
or P.AC.— ^ W.CB. CD 2 =0, 

^AC Txr CB 

W AC CD , . W CD 

% p" = ^c'C 7 D ; n = BC > "p = (XD' 

as found in Art. 143. 

193. Prop. 77«e principle of virtual velocities obtains in the 
single movable pulley with parallel cords. 




If the pulley A be raised to A', 
we shall have AA'=WW'=£PP', 
since each of the cords passing 
round Jhe pulley A must be short- 
ened by a length =WW. WW, 
the virtual velocity of W, is nega- 
tive. 

.-. P.PP'-W.WW'=0, 
or P.PF-W.1PF=0. 

W 
Hence -p-=2, as found in Art. 158. 



194. Prop. The principle of virtual velocities obtains in the 
first system of pulleys. 

In the figure of Art. 160 we see that, if W be raised through 
a space WW, each of the n cords at the lower block will be 
shortened by the same quantity, or that P will descend through 
a space n.WW. Hence the equation of virtual velocities 
P.PP'-W.WW'=0 becomes 

P.?i.WW'-W.WW'=0, or ~=n, 



as in Art. 160. 



PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 119 

195. Prop. The principle of virtual velocities obtains in the 
second system of pulleys. 

Referring to the figure in Art, 161, we see that, if P de- 
scends through the space PP', 

pp, 

the pulley a n would rise through a space =-77-, 

PP' 

pp, 

a n a a » 

" 3 2"~ 2 ' 

pp, 

a n a a a 

a 2 2 n_1 ' 

pp, 

" a j or the weight W " — -, 

And the equation of virtual velocities P.PP' — W. WW ; =0 
becomes 

pp, 

p.pp-w.— r =o, 
w 

or -p-=2", 

the same as in Art. 161, Cor. 2. 

196. Prop. Tfte principle of virtual velocities obtains in the 
third system of pulleys. 

Referring to figure of Art. 162, and designating the pulleys 
as in that article, we see that, if W be raised a space =WW, 
each cord will be shortened by a space equal to WW'. The 
highest movable pulley a n _ x will descend a distance =WW, 
The next pulley a„_ 2 will descend a distance =2. WW' by the 
descent of a n _ lt and a distance WW' by the elevation of W, or 
will descend on the whole (2+1) WW'. 

Similarly, the pulley <z„_ 3 will descend through 

{2(2 + l) + l }W.W'=(2 2 +2 + l)WW'. 

Proceeding in the same way, we find that pulley tf„_ ( „_3), or 
a 3 , will descend through the space 



120 



STATICS. 



(2"- 4 +2 n - 5 +, &c 2 + 1) WW, 

and a 2 through the space (2"- 3 +2"- I +,&c. . . . 2 + i)WW, 

and a, " " " (2"- 2 +2"- 3 +,&c 2 + l)WW, 

and P will descend through twice the last found space by the 
descent of the pulley a 19 and through the space WW by the 
elevation of the weight ; 

or PP=WW.{2(2"- 3 +2 n - 3 +, &c 2 + l) + l} 

= WW / .(2- 1 +2"- 2 +, &c 2 2 +2 + l) 

=WW'(2"-1). 
The equation of virtual velocities is 

P.PF-W.WW'-O, 
which becomes 

P.WW'(2"-1)-W.WW=0, 

w 



or 

as in Art. 162. 



=2 n -l 



197. Prop. The principle of virtual velocities obtains in the 
inclined plane. 

Let the force P make any an- 
gle e with the plane, a=z_BAC, 
a the first posilion of the body 
whose weight is W, a' the posi- 
tion of it after displacement. 

Drawing the perpendiculars 
av, a'u, we have —a'v=—aa' 
cos. e= the virtual velocity of 
P, and au—aa'. sin. a— the vir- 
tual velocity of W. 

By the equation of virtual velocities, 

or P.aa'. cos. e—W.aa'. sin. a=0. 

W sin. £ 

Hence -=r= , 

r sin. a 

as found in Art. 164. 




198. Prop. The principle of virtual velocities obtains in the 
wedge. 



PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 121 



Let 2P be the whole power, R 
and R the pressures perpendicu- 
lar to the faces of the wedge ABC, 
which produce equilibrium. Let 
the wedge be displaced to the posi- 
tion A'B C. The displacement of 
the point of application of Pis aa'= 
AA ; ; that of b, the point of applica- 
tion of R, is bb'=Am, a perpendicu- 
lar from A on A'B', and A/h=AA / 

. BAC 
Bin.— -. 

The equation of virtual velocities is 
P.«0'-R.&&'=O, 




or 



P.AA' 



x, * • BAC 
-R.AA'.sm.— — = 
lit 

p 
\ R= 



BAC' 

sin.—- 

as found in Art. 169. 

199. Prop. The principle of virtual velocities obtains in a lev- 
er of any form. 

Let ACB be the lever be- 
fore displacement, A'C'B' 
its position afterward. 
From A' draw A'v perpen- 
dicular to AP, and from B', 
B'u perpendicular to BQ, 
produced. Av is the vir- 
tual velocity of P, and Bu 
that of Q. Now when the displacements are indefinitely 
small, the circular arcs AA', BB' become straight lines, and 

Au=AA' cos. A'Au=AC.z_ACA'. cos. (PAC-90 ) 
= AC. sin. PAC.Z.ACA'; 

Bu=BC. sin. QBC.Z.BCB', 
and ^ACA' = z_BCB'. 

The equation of virtual velocities is 




122 



STATICS. 



or 



¥.Av-Q.Bu=0, 
P.AC. sin. PAC-Q.BC. sin. QBC 
P BC. sin. QBC 



as found in Art. 134. 



Q AC.sin.PAC 



200. Prop. The principle of virtual velocities obtains in the 
single movable pulley with cords inclined. 

Let A be the point where 
the cords produced would 
meet at the first position 
of the pulley, when P and 
W are the positions of the 
power and weight. 

Let P be displaced to 
P', when the weight is 
raised to W, or the point 
of meeting of the cord is 
raised to A'. Draw the 
circular arcs A'm, A'n with 

centers B and C. When the displacement is indefinitely small, 

the arcs A'?n, A'n become straight lines, and 

A?tz=AA' cos. BAA'=A?z, 




'PV'=A?n+An- 



BAC 

:2AA' cos. — -, WW'=AA'. 



The equation of virtual velocities is 

RPP'-W.WW'=0, 

which becomes 

BAC 



P.2AA'. cos. 



or 



W 
-p=2 cos. 



-W.AA' 
BAC 



0, 



as found in Art. 158. 

Scholium. In the preceding propositions, the expression 
P.PP'=W.WW, 



PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 123 

or pp/ - w , 

explains the principle that, " in using any machine, what we 
gain in power we lose in time." For, in order that W may be 
moved through any given space, we must have the space 
moved through by the power P, increased in the same ratio 
that P is diminished. 



CHAPTER VIII. 

FRICTION. 

201. The surfaces on which bodies pressed have hitherto 
been regarded as perfectly smooth, so that they offered no re- 
sistance to motion parallel to themselves, their only reaction 
being perpendicular. 

When rough surfaces are in contact, the motion, or tenden- 
cy to motion, parallel to the surfaces, is affected by the rough- 
ness, and the effect is called Friction. 

Friction may be divided into two kinds : sliding friction, 
when one rough surface slides on another, and rolling friction, 
when one rolls on the other. The former only will be consid- 
ered here, under the term Statical Friction. 

202. The laws of friction are determined by experiment. 

If the body A rest upon 

the perfectly smooth 

>-F plane BC, the smallest 

possible force applied to 

it will cause it to move. 
But if the body or the 
plane, or both, be rough, 
a force within certain 



0" 



A 



© 



limits of magnitude may 



be applied to it without causing motion. The greatest force 
which can be so applied to the body in the direction of the 
plane will measure the friction. 

Let W be this force, acting by a cord over a pulley on the 
body A, F being the opposing force of friction. Then F=W. 

203. The following laws of friction are deduced by this or 
some similar process. 

1°. The friction of the same body, or a body of the same ma- 
terial, when the weight is the same, is independent of the extent 



FRICTION. 125 

of the surfaces in contact, except in extreme cases, where the 
weight is very great compared with the surfaces in contact. 
Thus the friction of the body A will be the same whichever 
side rest on the plane, or whatever be the form within the ex- 
cepted limits. 

2°. The friction is proportional to the pressure on the plane, 
or the reaction of the plane, within moderate limits. If other 
weights, as m, be placed on A, W or F will vary as the whole 
reaction R of the plane. 

Cor. The friction is therefore some function of this pressure, 
and we may represent it by /xR. \i is called the coefficient of 

. F 

friction, and is equal to ^ , or the ratio of the friction to the re- 
action of the plane. 

204. Prop. The coefficient of fraction between two given sub- 
stances is equal to the tangent of the inclination of the plane 
formed of one of the substances, when the body formed of the oth- 
er is about to slide down it. 

Let the inclination a of the plane AC be increased till the 
body a is just on the point 
of sliding down it. The body 
a will then be in equilibrium 
from the normal reaction of 
the plane R, the friction jwR 
acting up the plane, and its 
weight W acting vertically 
downward. Y W 

Resolving parallel and perpendicular to the plane, we have 

juR — W. sin. a=0, 
R — W cos. a=0. 

Eliminating R, we have 

[i cos. a — sin. a=0, 
or jU^tan. a. 

205. Prop. To determine the limits of the ratio ofVtoW on 
an inclined plane, when friction acts up or down the plane. 




126 



STATICS. 



Let the power P make an 
angle e with the plane whose 
inclination is a, and W the 
weight of the body. 

1°. To determine the great- 
est value P can have without 
causing the body to move up 
the plane. In this case, the 
friction fiR, opposing the motion 
up the plane, will act down it, and, resolving parallel and per- 
pendicular to the plane, we have 

P. cos. e— juR— W sin. a=0, (a) 

P. sin. e+ R-Wcos. a=0. (b) 

Multiplying (b) by \i and adding to (a), we have 
P(cos. e+ju sin. e) — W. (sin. a+[i cos. a)=0, 




or 



P= 



W. (sin. a-\-\i cos. a) 



(60 



cos. £+i^ sin. £ 

2°. To determine the least value P can have without causing 
the body to move down the plane. In this case the friction 
will oppose the descent, and will therefore act up the plane. 
Hence equations (a) and (b) become 

P. cos. £+|uR— W sin. a=0, 
P. sin. £+ R — Wcos. a=0. 
Multiplying (b') by \i and subtracting, we find 
W. (sin. a— \i cos. a) 
cos. £— ^ sin. £ 
No motion will take place while the value of P is between 
these two, which are its limits. 

These two values of P may be combined so as to take the 
form 

W cos. edzfi sin. £ 
P — sin. a±(i cos. a 
in which the upper sign is to be taken when friction acts down 
the plane, and the lower when the friction acts up the plane. 

208. Prop. To determine the limits of the ratio of P to W in 
the screw, when friction acts assisting the power or the weight 



FRICTION. 



127 




Proceeding as in Art. 
.173, let ABC be the 
inclined plane formed 
by unwrapping one / 
revolution of the 
thread, the angle BAC 
= a ; let W= the 
whole weight sustain- 
ed by the screw, w= 
that part of it support- 
ed at a, Q= the whole 
force acting at the cir- 
cumference of the cyl- 
inder, ?-ED the radius of the cylinder, a=FD the lever at 
which the power acts, and q that part of Q, which supports w 

at a. Then Q=P-. 
r 

The forces which are in equilibrium at a are the weight id, 
the reaction R, the horizontal force q, and the friction j^R act- 
ing up or down the plane. 

Resolving parallel and perpendicular to the plane, we have 
q. cos. adbjuR — w. sin. a=0, (a) 

q sin. a— R-\-w cos. a=0. (b) 

Multiplying (b) by [i, and adding and subtracting, we have 
q(cos. a±{i sin. a) — w(s'm. az^.\i cos. a) = 0, 
q Q, sin. azpy, cos. a 



or 



w 
P 



W cos. adc/i sin. a 
r sin. azpfj- cos. a 



W a' cos. a±jLt sin. a 

The two values of this expression give the limits required, 

P 

and ^ may have any intermediate value. 

207. Cor. From the two preceding propositions it will ap- 
pear that, when we have obtained one of the limits, the other 
may be had by simply changing the sign of ft in the former. 



128 



STATICS. 



EXAMPLES. 

Ex. 1. A uniform straight beam rests on a rough cylinder 
of given radius ; required the greatest weight that can be sus- 
pended from one end of the beam without causing it to slide off. 
„2^ Let g be the center of gravity of the 
beam BC, whose length is 2d, and B^- 
=d since the beam is uniform, w= its 
weight, and W the weight suspended 
from B. Before the weight W was 
suspended from the beam the point g 
must have been at A. Let A' be the 
point of contact with the cylinder when 
the beam is on the point of sliding off, 
a the angle it then makes with the ho- 
rizon, and r= the radius of the cylinder. 
Resolving parallel and perpendicular to the beam, we have 




1-iR — w. sin. a — W sin. o,=0, 
R — w cos. fl-Wcos. a=0 ; 



(a) 



whenc: 



tan. a. 



Taking the moments about A', we have 



=0, 



W.BA' cos. a—w.A'g. cos. cl- 
oy W.(Bg—A'g) — w.A'g=0. 

But A'g= arc A'A= radius X angle AOA'=r.«. 

.-. W.(d—ra)—w.ra=0, 



or 



W: 



w.ra w.rtan. 



-, the weight required. 



d—ra d— rtan." ~\i 

Ex. 2. A ladder rests with one end on a rough horizontal 
plane, and the other on a rough vertical wall ; given Z= its 
length, d= the distance of its center of gravity from its lower 
end, i-i and fi'= the coefficients of friction on the horizontal and 
vertical planes respectively ; required its inclination 6 when 
on the point of sliding down. 

Let AB be the ladder and the forces acting upon it, as rep- 
resented in the figure. 



EXAMPLES ON CHAPTERS VI., VII., AND VIII. 129 

Resolving horizontally and 
vertically, we have 

Rz-^R^O, (a) 

R+p'R'-w=0. (b) 

Multiplying (a) by p', and sub- 
tracting from (b), we find 

w = R(l+^0» 
and from (a), R'=juR. 

Taking moments about A, we 
get 

w.d. cos. d—R'.L sin. d—p'R'J. cos. 0=0, 
w.d-fi'R'l 




or 



tan. 6= 



R7 



Substituting the values of w and R', 

d.(l+w>)-lwi' 



tan. 6= 



\d 



If the center of gravity of the ladder be at the middle point, 
l=2d, and 

l-fifi' 



tan.0=- 



2f* 



208. EXAMPLES ON CHAPTERS VI., VII., AND VIII. 

Ex. 1. A beam 30 feet long balances on a prop -^ of its 
length from the thicker end; but when a weight of 10 lbs. is 
suspended from the other end, the prop must be moved 2 feet 
toward it to maintain the equilibrium. Required the weight 
of the beam. 

Ex. 2. The forces P and Q, act at arms a and b respectively 
of a straight lever, which rests on a fixed point to which it is 
not attached. When P and Q, make angles a and (3 with the 
lever, required the conditions of equilibrium. 

Ex. 3. A uniform beam is sustained by three persons, one 
at one end, and the other two by a hand-spike placed at some 

I 



130 STATICS. 

point beneath it. At what point must the hand-spike be placed 
that each person may sustain one third of the weight? 

Ex. 4. A Roman steelyard, whose weight is 10 lbs., has its 
center of gravity 2 inches from the fulcrum, and the weight to 
be determined is supported by a pan placed at a distance of 3 
inches on the other side. Find the respective distances from 
the fulcrum at which the constant weight of 5 lbs. must be 
placed, in order to balance 10, 20, 30, &c.,lbs. placed success- 
ively in the pan. 

Ex. 5. Find the ratio of the power to the pressure in the 
common vice. 

Ex. 6. Find the ratio of the power to the pressure in the 
screw by the principle of virtual velocities. 

Ex. 7. An isosceles triangle, whose base is to one of its 
equal sides as 1 : V 7, is placed with its base on an inclined 
plane ; and it is found that, when the body begins to slide, it 
also begins to roll over. Find the coefficient of friction. 

Ex. 8. A ladder rests against a vertical wall, to which it is 
inclined at an angle of 45° ; the coefficients of friction of the 
wall and of the horizontal plane being respectively ^ and \, 
and the center of gravity of the ladder being at its middle 
round. A man whose weight is half the weight of the ladder 
ascends it. Find to what height he will go before the ladder 
begins to slide. 

Ex. 9. In a uniform lever of the second kind, which weighs 
2 ounces per inch, required the length of the lever, in order 
that the power may be the least possible when in equilibrium 
with a weight of 48 ounces placed at a djstance of three inches 
from the fulcrum. 

Ex. 10. Two given weights, P and Q, are suspended from 
two given points in the circumference of the wheel, a being 
the angle made by the radii drawn to the points of suspension. 
Required the angle which the lower radius makes with the 
vertical when the weights cause the greatest pressure on the 
axle. 



DYNAMICS. 



INTRODUCTION.— DEFINITIONS. 

209. In Statics we have investigated the relations of the in- 
tensities and directions of forces necessary to produce equilib- 
rium, this result being entirely independent of the time during 
which the forces act. 

In Dynamics, forces are regarded as producing motion or 
change of motion in bodies, and these effects must obviously 
depend on the duration of the action of the forces. In dy- 
namics, therefore, time becomes an element in our investiga- 
tions. 

210. Motion is the transit of a material point or body from 
one position to another in space. 

211. The absolute motion of a body is its transit from one 
fixed point in space to another. 

212. Relative motion is a change of distance from a point 
which is itself in motion. 

All motions are relative in any practical view which we 
can take of them, since we have no means of determining the 
absolute rest of any point in space. 

213. The velocity of a body is its rate of motion. It may be 
uniform or variable. 

The velocity of a body is uniform when it passes over equal 
spaces in equal times, and is measured by the space passed over 
in the unit of time. 

Let v be the velocity, or space passed over in one second, 
then the space described in two seconds will be 2v, in three 
seconds Sv, and so on ; and if s be the space described from 
the commencement of motion, and t the number of seconds, 
also reckoned from the commencement of motion, then 



132 DYNAMICS. 

s=vt. [I.] 

The units of time and space are arbitrary. It is usual to take 
one second for the unit of time, and one foot for the unit of 
space. When no mention is made of different units, these will 
be understood. 

214. Variable velocity is that which continually increases or 
decreases, so as to be the same at no two successive instants. 
To find a measure of the velocity of a point or body so moving, 
let us assume, at first, that the velocity which the body has at 
the end of t x seconds is uniform from ^seconds to t 2 seconds, 
a very small interval, the space passed over at the end of t x 
seconds being s l9 and the space passed over at the end of t 2 
seconds being s 2 . Then, using the symbol A to signify finite 
difference, by [I.] we have 

s 2 -s x As 



t 2 -t, At 
If, however, there is no time, however small, during which the 
velocity is uniform, then the smaller we take A£, and conse- 
quently As, the more nearly 

-£• m 

That is, the velocity, when variable, is measured by the limit 
of the ratio of the space described to the time of describing it. 

215. Relative velocity is the velocity with which two bodies 
approach or recede from each other. 

216. Matter at any given moment must be in one of the two 
states, motion or rest. The inertia of matter is the entire ab- 
sence of power in itself to change this state. It implies equal- 
ly a disability, when in motion, to change its rate or its direc- 
tion. Hence 

A body, when not acted on by any external forces, if at rest, 
will remain so, or, if in motion, will continue to move in a straight 
line and with a uniform velocity. 

This is called the first law of motion. 

217. It is a consequence of the inertia of matter, that when 
a force is applied to a body to move it, each of its particles 
opposes a resistance to motion in directions parallel but oppo- 



DEFINITIONS. 133 

site to the direction of the applied force. The center of these 
parallel forces {Art. 44) of resistance is called the center of in- 
ertia of the body. It is the same point which, in statics, was 
called the center of gravity of the body in reference to the 
force which was there supposed to act on the body. 

The sum of these parallel resistances, or their resultant 
{Art. 43), is obviously proportional to the number of particles 
in the body or to the whole mass. Hence the inertia of a 
body is a surer test of the quantity of matter or mass of a 
body than its weight is ; for the latter {Arts. 84 and 88) varies 
by a change of position on the earth, while the former is al- 
ways the same. 

218. The path of a body is the line, straight or curved, 
which its center of inertia describes when it passes from one 
point to another in space. 

219. A body is said to be free or move freely when its path 
depends on the action of the impressed forces only. Its mo- 
tion is said to be constrained when its path is limited to a given 
line, straight or curved, or limited to a given surface. 

220. An impulsive force is one which acts instantaneously 
and without sensible duration. 

221. An incessant force is one which acts without intermis- 
sion. If a material point move from rest by the action of an 
incessant force, its rate of motion or velocity must continually 
increase. The amount of this increase, or the increment of the 
velocity in the unit of time, will obviously be greater or less as 
the intensity of the force is greater or less. This increment 
of the velocity in one second is therefore the measure of the 
intensity of the force. 

222. A constant force is an incessant force whose intensity 
is at all times the same. If a material point move by the ac- 
tion of a constant force, the increments of the velocity in each 
successive unit of time must all be equal, and each increment 
will be a measure of the force. If, therefore, we put <p for the 
increment of velocity, or the velocity generated by the force 
in one second, <b will represent the force. The increments be- 
ing all equal, the velocity generated in two seconds will be 



134 DYNAMICS. 

2(j>, in three seconds 30, and so on. Hence, if v be the velocity 
generated in t seconds, 

V = (f)t. [III.] 

223. A valuable force is an incessant force whose intensity 
either increases or decreases, so as to be the same at no two 
successive instants. To find an expression for a variable 
force, let us assume it to be constant from the end of the time 
t l , when the velocity is v lt to the end of the time t a , when the 
velocity is v 2 . Then, by [III.], we have 

v„ — v, Au 
r t 2 —t l M 

If, however, there be no interval during which the force is 
constant, then the smaller we take &t, and consequently Au, 
the more nearly will 

Av 
0=—. [IV.] 

That is, a variable force is measured by the limit of the ratio 
of the velocity caused by it to the time of causing it. 

224. The momentum of a body is its quantity of motion, and 
is measured by the product of the mass of the body by its veloc- 
ity. For the motion of a single particle is its velocity, and the 
motion of any number of particles, having the same velocity, 
is obviously as much greater as the number of particles is 
greater. Hence the whole motion is equal to the whole num- 
ber of particles in the body, or its mass, multiplied by their 
common velocity. Or, if Q be the quantity of motion of a 
body, M its mass, and V its velocity, 

Q=MV. [V.] 

The mass, multiplied by the square of the velocity, is called the 
vis viva, or the living force of a body. 

225. In estimating the effects of incessant forces, we have 
considered only the acceleration or velocity which each force 
will produce when acting on a free material point or a unit of 
mass. When so measured, they are called accelerating forces. 
If the mass moved differs from that which we have called the 
unit of mass, and it is taken into consideration in estimating 



DEFINITIONS. 135 

the effects of the forces, they are then called moving forces. 
Let <p be the acceleration, or velocity generated in a unit of 
time, M the number of units of mass ; then $, the moving 
force, will be measured by the quantity of motion generated 
in the unit of time, or 

*=0M. [VI.] 

Hence (f>= ™r, or the accelerating force, is equal to the moving 

force divided by the mass. 

226. If a body already in motion be acted on by a force in 
the direction in which the body moves, the superadded motion 
is just the same as that which would have been produced in 
the body if at rest when the force began to act-. 

If the force act in a direction different from that in which 
the body moves, the new motion produced by the force, esti- 
mated in the direction of the force, will be the same as if the 
body had been at rest. 

If the force act in a direction opposite to that in which the 
body moves, the motion destroyed in the body is equal to that 
which the force would produce in the body if at rest when the 
force began to act. 

These facts are consequences of the inertia of matter, and 
will receive additional illustration in the sequel. They are 
embodied in the following enunciation, called the second law 
of motion: 

All motion or change of motion in a body is proportional to 
the force impressed and in the direction of that force. 

227. When one body impinges on another at rest or in mo- 
tion, the quantity of motion, or momentum of the two bodies 
after impact, is the same as before impact. For matter being 
incapable of originating motion, can not add to the motion of 
other matter, or take from it except by imparting its own mo- 
tion. Hence whatever motion the second body receives in 
the direction of the striking body, just so much must be lost by 
the striking body. 

This fact is usually enunciated as follows, and is called the 
third law of motion : 

Action and reaction are equal and in opposite directions. 



CHAPTER I. 

UNIFORM MOTION. 

228. In considering the effect of an impulsive force, we shall 
suppose the force applied, at the center of inertia (Art. 217). 
When so applied the parallel forces of resistance of all the 
particles situated on opposite sides of the direction of the 
force will balance each other, and the body will not rotate, 
but all its particles will describe parallel lines with a common 
velocity. 

229. Prop. To find the general equation of uniform motion. 
X m Let OB be the path of the body (Art. 218), v its veloci- 
ty (Art. 213), and t the time of its motion in seconds. 
Let O be the origin, or point from which we estimate the 
successive positions of the body, s the distance of the body 
from O at the end of the time t, and OA=s 1 its distance 
from the origin O at the commencement of the time. 

When the body is at B, we have [I.] 

AB=^. 
But OB-OA+AB, 

B or s=s,+Vt, (36) 

which is the general equation of uniform motion. 

If the body, instead of receding from the origin O, approach 
it, then v will be negative ; or if the body begin to move from 
A', then s T will be negative. 

Cor. 1. From (36) we obtain 

s—s, AB 

or the velocity of a body is equal to the constant ratio of the 
space described to the time of describing it. 

Cor. 2. If we estimate the position of the body from the 



UNIFORM MOTION. 137 

point where the body is when 2=0, or suppose the space and 
time to commence together, then s l =0, and 

s=vt. (37) 

230. Prop. If two bodies move during the same time, their 
velocities will be proportional to the spaces described by them 
respectively. 

Taking the points of departure for the origin of spaces, we 
have from (37) 

s=vt and s'=v't'. 
.*. s : s'=vt : v't'. 
And, since t=t', s : s'=v : v'. 

231. Prop. If the velocities of two bodies are equal, the spaces 
described are proportional to the times. 

As before, s : s'—vt: v't' ; 

and if v=v', s : s'= t : t'. 

232. Prop. If the spaces described by two bodies are equal, 
their velocities are reciprocally proportional to the times. 

For, since s : s'=vt : v't, 

ifs=s', vt=v't'. 

.'. v : v'=t' : t. 

233. Prop. An impulsive force is measured by the momentum 
it can produce in any mass. 

If v be the velocity produced by the force F x in a body con- 
taining M units of mass, by [V.] the momentum will be Mv. 
Now if the mass M move from rest by the action of the force, 
all the motion it receives is the effect of the force. And, ad- 
mitting the principle that effects are proportional to their 
causes, if A be the constant ratio of the force to the momentum, 
then 

F^X.Mv. 

But the unit of force being arbitrary, we may assume it to be 
A. Hence, putting F for the number of units of force, or the 
ratio of F T to A, we have 

F=^=Mv. (38) 

Cor. If M=l, F l =X.v, in which A is constant. Hence the 
force is proportional to the velocity produced in the unit of mass. 



138 



DYNAMICS. 



Schol. Since we know nothing of the nature of forces, we 
can not determine their effects a priori. The foregoing propo- 
sition ought, therefore, to be regarded as depending ultimately 
on observation and experiment. The fact that on the earth, 
which is subject to the double motion of rotation and transla- 
tion, forces are found to produce precisely the same effects as 
if the earth were at rest, is a confirmation of its truth ; also 
that a pendulum performs its vibrations in the same time, what- 
ever be the direction of these vibrations in reference to an east 
and west line. 

234. If any number of forces act upon a body in the same 
direction, the velocity imparted to the body will be equal to the 
sum of the velocities imparted by each. 

Let F, F', F" . . . be any number of forces acting upon a 
body, v, v', v" . . . the velocities imparted by each respective- 
ly, </> the resultant force, and u the velocity due to 0. 

Then F=Xv, F'=W, Y"=Xo", &c. Since the forces are 
conspiring, their resultant will be equal to their sum. 

.-. F-!-F'+F"+ &c, =<}>=X(v+v' + v"+ &c). 
But 0=Aw. 
Hence u—v-\-v> l ~\-v"-\- &c. 

235. Prop. If the adjacent sides of a parallelogram represent 
in magnitude and direction the velocities which two forces, by 
their separate action, would respectively produce in a body, the 
diagonal of the parallelogram will represent the actual velocity 
produced by their joint action. 

Let AB and AC represent the velocities which the forces 

P and Q respectively im- 
part to a body placed at 
A, or, in other words, the 
spaces over which the 
body would pass in a unit 
of time by the separate ac- 
tion of the forces, then AD 
will represent the velocity 
produced by their result- 
ant, or the space over 




UNIFORM MOTION. 



139 



which the body will pass in the unit of time by the joint action 
of the forces. 

We may assume AB to represent the force P ; then, by Art. 
233, Cor., 

P : Q=AB : AC, 
and AC will represent the force Q. Also, Art. 21, AD will 
be the resultant R of P and Q. Let x be the velocity due to 
the resultant R, then (Art. 233, Cor.) 
P : R = AB : x. 
But P:R=AB:AD. 

.-. x=KD, 
and AD being the direction of the resultant, the velocity due 
to it will have the same direction. 

The two preceding propositions illustrate the second law 
of motion (Art. 226). 

236. Prop. The velocity of a body being given, to find the com- 
ponent velocities in any directions at right angles to each other ; 
and two component velocities at right angles to each other being 
given, to find the resultant velocity. 

Let AD represent the velocity v of 
the body in direction and magnitude, 
AX and AY the rectangular directions 
in which the components are required, 
and the angle which AD makes with 
AX. Completing the rectangle, AB 
will represent the velocity a in AX, 
and AC the velocity b in AY. 

A 

Then AB=AD. cos. 0, AC = AD. sin. 6, 

or a=v. cos. 6, b=v.s'm.O. (39) 

Again, if a and b represent the given component velocities 
in AX and AY, to find the resultant velocity v we have 

v = Va 2 +b\ (40) 

b 




and 

as required. 



tan. 6=- 



(41) 



140 DYNAMICS. 

Cor. Hence velocities may be compounded and resolved 
like forces in statics. 

237. Prop. Two bodies, A and B, describe the same path with 
the velocities v and v', and at the commencement of their- motion 
are at a distance a from each other : to find the time t when they 
will be at a distance b from each other and the position of each 
at the end of that time. 

Taking the position of A when t=0 for the origin of spaces, 

the equations of their motions are, by (37) and (36), 

s=vt, (a) 

and s'=a + v't'. (b) 

By the conditions of the question, t—t'. 

Also s—s'=b or s' — s=b, .*. s— s'=±b. 

Subtracting (b) from (a), and putting t=t', we have 

s— s'=vt— a— v't=±b. 

a±b , x 

.'. t= ? (42) 

v—v' v 

Also, from (a) s=vt=v. _ , , (43) 

and from (b) s'=a+v't=a+v'. :== — . (44) 

w v — v v — v 

Cor. 1. If the bodies move in opposite directions, v' will be 

negative, and 

a±b a±b va^iv'b 

t=z ~r~^ s=v -~^^> s '= — r~r- 

v-\-v' v+v v-\-v 

Cor. 2. If the time when the bodies are together be required, 
then b=0, and 

a va v'a 
t= :, s= =s', s'—a= :. 

v—v' v — v v—v 

238. Prop. When two bodies A and B move in the circum- 
ference of a circle with uniform velocities, to determine the cir- 
cumstances of their motion. 

Let v and v' be their velocities, c the circumference of the 
circle, and a the distance apart at the commencement of the 
time. Then, putting b=0 in (42), we have 



UNIFORM MOTION. 141 



t ,== j = the time of their first meeting. 



v—v 

a-\-c 



" second " 



a+2c . . , 

t = r = " " third 



a-\-(n— l)c 

v—v' 

also, s=u. ; — = space described by A, 

v — v 

v'a + v(n—l)c _ 

and s-a= ^ — — = " " B. 

v — v' 

Cor. The interval between two successive conjunctions is 

**-'.=''=^- ( 45 > 



239. EXAMPLES. 

Ex. 1. If an iron rod have one end against the sun and the 
other resting on the earth, the distance of the sun from the 
earth being 95,125,000 miles, in what time will a blow applied 
to the end on the earth be felt by the sun, the velocity of an 
impulse in iron being 11,865 feet per second ? 

Ans. 490 days. 

Ex. 2. When the earth is in that part of its orbit nearest to 
Jupiter, an eclipse of one of Jupiter's satellites is seen 16 min- 
utes 36 seconds sooner than it would be if the earth were in 
that part of its orbit most remote from Jupiter. The radius 
of the earth's orbit being 95,125,000 miles, what is the veloci- 
ty of light ? 

Ex. 3. The star 61 Cygni is ascertained to be 56,319,996,- 
600,000 miles distant from us ; were its light suddenly extin- 
guished, in what time would the intelligence reach us, the 
velocity of light being 191,000 miles? 

Ex. 4. Suppose 964 tons of ice to be floating directly to the 
east at sunrise on the 21st of March, with a velocity of 12 feet 



142 



DYNAMICS. 



per minute, how many grains of light from the sun would be 
sufficient to stop it ? 

Ex. 5. A train of cars moving with a velocity of 20 miles 
an hour, had been gone three hours, when a locomotive was 
dispatched in pursuit, with a velocity of 25 miles an hour ; in 
what time did the latter overtake the former? 

Ex. 6. Had the trains in Ex. 5 started together and moved 
in opposite directions around the earth, 24,840 miles, in what 
time would they meet ? 

Ex. 1. Suppose it to be 12 minutes past noon by a clock; 
in how long a time will the hour and minute hands of the 
clock be together ? 

Ex. 8. The daily motion of Mercury in his orbit is 4°.09239 ; 
that of Venus 1°.60216; that of the earth 0°.98563 : what are 
the intervals between the epochs at which Mercury and Venus 
respectively will be in the same direction from the sun as the 
earth ? 

Ex. 9. A man being caught in a shower in which the rain 
fell vertically, ran with a velocity of 12 feet per second. He 
found that the drops struck him in the face, and estimated that 
the apparent direction of the drops made an angle of 10° with 
the vertical line. What was the velocity of the drops ? 

Ans. 68 feet. 

Ex. 10. When the path of the earth in its orbit is perpen- 
dicular to a line drawn from a star to the earth, the path of the 
light, from the star appears to make an angle of 20^,445 with 
the perpendicular to the path of the earth. The velocity of 
the earth being 68,180 miles per hour, what is the velocity 
of light? 



CHAPTER II. 

IMPACT OF BODIES. 

240. Def. When two bodies in motion impinge, if their cen- 
ters of inertia move in the same straight line perpendicular to 
a plane tangent to the bodies at their point of contact, the im- 
pact is said to be direct and central. 

If the straight line described by the center of inertia of one 
of the bodies is not perpendicular to the tangent plane, the im- 
pact is said to be oblique. 

In the cases discussed the bodies will be supposed, spherical 
and of uniform density. 

241. Def. When the bodies impinge, they exert a mutual 
but varying pressure during the interval between contact and 
separation, an interval of time which is generally very short, 
and we suppose them to suffer a degree of compression, by 
which, during a portion of this interval, their centers will ap- 
proach each other, and during the remaining portion will re- 
cede by the action of an internal force tending to restore them 
to their original form. The force urging the approach of their 
centers is called the force of compression ; the opposing force 
causing them to separate again is called the force of restitution 
or elasticity. The ratio of the force of restitution to that of 
compression is called the modulus of elasticity. 

When this ratio is unity, or the force of restitution is equal 
to that of compression, the bodies are perfectly elastic ; when 
it is zero, or the force of restitution is nothing, they are inelastic. 
If the value of the ratio is intermediate between zero and unity, 
the bodies are imperfectly elastic. 

242. Def. If the bodies suffer no compression, they are call- 
ed hard; if, when compressed, they exert no force to recover 
their original form, they are called soft. 

There are no known bodies either perfectly elastic or per- 



144 DYNAMICS. 

fectly inelastic, but these states may be considered as limits to 
the various degrees of elasticity presented in nature. 

243. Prop. To determine the velocity of two inelastic bodies 
after direct impact. 

Since the bodies are inelastic, the force of restitution is zero, 
and the bodies will move on together with a common velocity. 

1°. Let m x and m 2 be the masses or bodies moving in the 

© mo same direction with the velocities 
>■ > Vj and v 2 (v,>u 2 ), and v their 

common velocity after impact. 
If F and F' are the forces which impress on the bodies their 
respective velocities, then (38) 

F=m 1 v l , and F'=m 2 v 2 , 
and their resultant, F-\-F' = m 1 v 1 -\-m 2 v 2 . 

After impact the bodies move on together as one mass, and 
its momentum (m l +m 2 )v must be a measure of the force 
F+F'. 

.'. (m 1 +m 2 )v=m 1 v l -\-m 2 v 2 , 

m.v,-\-m„v<, . . 

or U== _±4J_ — ?_2_. (46) 

m 1 -\-m 2 

2°. If the bodies move in opposite directions, the resultant 
of the forces will be F— F'. 

.*. F— F'=m l v 1 — m 2 v 2 = (m 1 +m 2 )v, 

»= "■'■-"■'■ . (47) 

The same result will be obtained by changing the sign of 
v 2 in (46). 

Hence the velocity of two inelastic bodies after impact is 
equal to the algebraic sum of their momenta before impact, di- 
vided by the sum of their masses. 

Cor. 1. If two bodies move in opposite directions, with veloci- 
ties reciprocally proportional to their masses, they will rest after 
impact. 

For m 1 : m 2 =v 2 : u, 

gives m 1 v 1 =m 2 v 2 , 

which, substituted in (47), gives v=0. 



IMPACT OF BODIES. 145 

Cor. 2. If m 2 is at rest before impact, v 2 =0, and 



m 1 -\-m 2 ' 
and if, at the same time, the masses are equal, 

Cor. 3. If the masses are equal, and move in the same di- 
rection, 

if in opposite directions, 

2 

244. Prop. In the impact of inelastic bodies there is a loss of 
living force, and this loss is equal to the sum of the living forces 
due to the velocities lost and gained by the bodies respectively. 

For the living force before impact = m l v 2 l +m 2 v 2 2 (A?~L224), 
and " " " after " = (m l -\-m 2 )v\ 

.*. the loss of living force by impact is 

myV 2 +m.j) 2 2 — (m 1 +m 2 )v 2 =m 1 v 2 -\-m 2 v 2 2 — 2(m l +m 2 )v i +(m 1 +m 2 )v z 
(46), =m i v 2 +m 2 v 2 2 — 2(m 1 v 1 +m 2 v 2 )v+(m 1 -\-m 2 )v- 

=m 1 (v i —vy+m. 2 (v. 2 —vy, 
which is necessarily positive, and in which v, — v is the veloci- 
ty lost by m l9 and v— v 2 the velocity gained by m 2 . 

From this proposition it appears that in machinery made of 
inelastic materials all abrupt changes of motion are attended 
with a loss of living force, by which loss the efficiency of the 
machinery is impaired. 

245. Prop. To find the velocities of two imperfectly elastic 
bodies after direct impact. 

Let m x and m 2 be the masses, v t and v 2 their velocities be- 
fore impact, and v' lt v' 2 their velocities after impact. 

The bodies being elastic, will suffer compression. Let v be 
their common velocity at the instant of greatest compression, 
or when the distance between their centers is least. Then the 
velocity lost by m l at this instant will be »,.— v. 

K 



146 



DYNAMICS. 



Let e be the modulus of elasticity, or the ratio of the force 
of restitution to that of compression. Since these forces are 
proportional to the velocities they generate or destroy in the 
same mass, the velocity destroyed in m , by the force of resti- 
tution will be e(v l — v). 

Hence the whole velocity lost by m l will be 
v 1 —v + e(v 1 —v) = (l+e) (v^—v). 
This, subtracted from the velocity of m l before impact, will 
give its velocity after impact, or 

v' l = v 1 — (l+s) (v l —v)=v—e(v 1 —v). (a) 

In like manner, the velocity gained by m 2 during compres- 
sion will be v — v 2 , and the velocity gained by the force of 
restitution e(v—v 2 ). 

Hence the whole velocity gained by m 2 will be (1+e) 
(v—v 2 ). This, added to the velocity before impact, will give 
the velocity after impact, or 

v 2= :v 2+( l + e ) (v — v 2 )=v+e(v—v 2 ). (b) 

Substituting in (a) and (b) the value of v (46), and reducing, 
v , = m,v,+m 2 v 2 m 2 e(v^-v 2 ) ^ 
1 m l J rm 2 m 1 +m 2 ^ 

As in Art 243, if the bodies move in opposite directions, we 
must change the sign of v 2 , or, if one of them be at rest before 
impact, make v 2 =0. Also, if we put e=l, the results will be 
those for perfectly elastic bodies, or make e=0, the result will 
be that for inelastic bodies. 

Cor. 1. If the bodies are perfectly elastic, their relative 
velocities before and after impact are the same. For, making 
e=l in (a) and (b), and subtracting the latter from the former, 
we have 

v'-v'=v 2 -v x . (50) 

Cor. 2. In the impact of bodies no motion is lost. 

For, multiplying (48) by m lf and (49) by m 2 , adding and 
reducing, we have 

m 1 v' l +m 2 v' 2 =m 1 v 1 +m 2 v 2 , (51) 



IMPACT OF BODIES. 147 

in which the first member is the sum of their momenta after 
impact, and the second member the sum of their momenta be- 
fore impact. 

Cor. 3. If the bodies are perfectly elastic and equal, they 
will interchange velocities by impact. 

For, making m 1 =m 2 and e=l in (48) and (49), we have 

v \=i( v i+ v 2)-i( v i- v 2 )= v 2 
v '2 = i( v i J r v 2 )+i( v i— v 2) =:iv i' 
Cor. 4. The velocity which one body communicates to an- 
other at rest, when perfectly elastic, is equal to twice the ve- 
locity of the former divided by one plus the ratio of the masses. 
Making in (49) m 2 =rm 1 , v 2 =0, and e=l, we obtain 

v * r+r 

246. Prop. When, in a series of n perfectly elastic bodies 
whose masses are in geometrical progression, the first impinges 
directly against the second at rest, the second against the third, 
and so on, to find the velocity of the nth body. 

Let m, ?-m, r 2 m .... r n ~ l m be the bodies, and v the velocity 

of m. By Cor. 4, Art. 245, 

2v 
the velocity of the second will be — — -, 

2v 1 2 2 v 

third " 2. 



1+r'l+r (1+r) 2 ' 

t , 2*v 1 2 5 v 

fourth " 2.- t 



*(l+r) 2 'l+r (1+r) 3 ' 



?ith 



(1+r)"- 

Cor. 1. The momentum of the nth body is 
2 n ~'v ( 2r Y~ l 

(1+r) l \l+r/ 

Cor. 2. If the bodies are equal, r==l, and the velocity of the 
last equals v. the velocity of the first. If all are in contact ex- 
cept the first before impact, all except the last will remain in 
contact after impact, and the last will move off with the ve- 
locity of the first. 



148 DYNAMICS. 

247. Prop. The motion of the common center of gravity of 
two bodies after direct impact is the same as before impact. 

Let v be the velocity of the common center of gravity of the 
bodies m 1 and m 2 , moving in the same direction with veloci- 
ties «j and v 2 respectively before impact, and v' the velocity 
of their common center of gravity after impact. Let x 19 x 2 , 
x be the distances respectively of the centers of gravity of m l 
and m 2 , and of their common center of gravity from any fixed 
point in their line of motion at any instant ; x' l9 x' 2 , x' the same 
quantities after an interval t, so that (36) 
x' 1 =x l +v 1 t, 
x' 2 =x 2 +v 2 t t 
x 1 =x +vt. 
By (29), Art. 105, 

(m 1 +m 2 )x =m 1 x l +m 2 x 2i (a) 

(m 1 +m 2 )x'=m 1 x' 1 +m 2 x 2 . (b) 

Substituting in (b) the values of x', x' 2 , x\ above, and subtract- 
ing (a) from the result, we have 

m 1 +m 2 v 

which is the same as the velocity of two inelastic bodies after 
impact, and therefore equal to the velocity of their common 
center of gravity after impact, since the masses move on to- 
gether. 

If the bodies are elastic, and v\ , v' 2 are their velocities after 
impact, then 

x' 1 =x l -\-v' 1 t, 

X 2 — X 2 ~i " 2 ' 

x' =x +v't; 
from which we deduce, as before, 

m x +m 2 v 

for the velocity of the common center of gravity of elastic 
bodies after impact. Now the denominators of (c) and (d) are 
the same, and by (51) the numerators are equal ; 

.*. v=v' 9 



IMPACT OF BODIES. 



149 



or the velocity of the common center of gravity is unchanged 
by impact. 

248. Schol. This proposition is only a particular case of a 
general principle in Mechanics, denominated the conservation 
of the motion of the center of gravity. The principle consists 
in this, that the mutual action of the several bodies or parts of 
a system upon each other can produce no change in the mo- 
tion of the center of gravity of the entire system. 

249. Def. If a body impinge on a surface, the angle which 
its path, before impact, makes with the perpendicular to the 
surface at the point of impact is called the angle of incidence, 
and the angle which its path, after impact, makes with the 
same perpendicular is called the angle of reflection. 

250. Prop. To determine the motion of a smooth inelastic 
body after oblique impact upon a smooth, hard, and fixed plane. 

Let the body m impinge on the plane 
AC at B, with the velocity v, making 
the angle of incidence mBN=0. 

The component of the velocity v 
parallel to the plane (39) is v. sin. ; 
and this velocity will not be changed 
by impact, since the body and plane are smooth. The com- 
ponent of v perpendicular to the plane, viz., v. cos. 0, will be 
destroyed by the plane, and, since the body and plane are in- 
elastic, there will be no vertical velocity after impact. Hence 
the body will slide along the plane with the velocity v. sin. 0. 

251. Prop. To determine the motion of an elastic body after 
oblique impact upon a smooth, hard, and fixed plane. 

Let the body m impinge on the plane 
RS at B, with the velocity v, making the 
angle of incidence PBN=0. Let PB rep- 
resent the velocity v before impact. Draw 
PR and PN perpendicular and parallel to 
the plane RS. PN=RB=u. sin.0 is the 
component of the velocity parallel to the plane, and is not af- 
fected by the impact. PR=NB=t>. cos. 0, the component of 
the velocity perpendicular to the plane, will be destroyed by 





150 



DYNAMICS. 



the plane. But the body being elastic, the force of restitution 
will give it a velocity s.v. cos. 6 in the direction BN. Take 
BM=£.d. cos. 6, BS=v. sin. 6, and, completing the parallelo- 
gram, draw BQ,. BQ,^' is the direction, and BQ, the velocity 
of the body after impact. Now 

BQ 2 =V 2 =BM 2 +MQ 2 

=£.V 2 cos. 2 0+v 2 sin. 2 0. 



also, 



BQ,=v'=vVe. 2 cos. 2 0+sin. 2 6 ; 
MQ_ PN __ tan.fl 
MB~e.NB~ s ' 



tan. 6' 



Cor. If e=l, 6=6', and v=v', or if the body be perfectly 
elastic, the angle of incidence equals the angle of reflection, 
and the velocity is the same after as before impact. If e=0, 
tan.0' = cz>, and 6'=90°. 

252. Prop. To determine the direction in which a body of 
given elasticity must be projected, in order that after reflection 
from a given plane it may pass through a given point. 

Let MB be the given plane, A the 
position of the body before projection, 
D the point through which it is re- 
quired to pass after reflection, and e 
the modulus of elasticity of the body, 
supposed known. 

Draw AMC perpendicular to the 
plane, and take AM : MC = 1 : e 
Draw C D cutting the plane in B, and join A B. AB is the 
direction in which the body must be projected, and AB, BD 




will be its path. 
Since 


e.AM=MC, 




MB MB 




e .AM~MC 


or 


tan. MAB ^ 

— =tan. MCB ; 

e 




tan. 6 

—tan. 6' 



or 



as in Art. 251, 



IMPACT OF BODIES. 



151 



253. Prop. The modulus of elasticity is equal to the ratio of 
the relative velocities of the bodies after and before impact. 
For, eliminating v from (a) and {b) (Art 245), we have 



(52) 



254. Schol. Bodies suspended by fine cords, and made to 
oscillate like a pendulum, acquire velocities at the lowest point 
proportional to the chords of the arcs of 
descent, and will rise through arcs whose 
chords are proportional to the velocities 
impressed upon them at the lowest point 
of the arcs. Also, if the arcs be small, 
the times of descent will be equal, so 
that bodies descending through arcs of 
different lengths will impinge at the 
lowest points. If, therefore, two spher- 
ical bodies of the same material be made to descend, in the 
manner described, through arcs of given length, and the arcs 
through which they rise after impact be measured, the veloci- 
ties v l9 v a , u',, v' 2 will be known, and these, substituted in (52), 
will give e the modulus of elasticity. 

The following table exhibits the results of experiments, per- 
fect elasticity being unity : 



(T^ 



2 10X2 



Substances. 


Moduli. 


Substances. 


Moduli. 


Glass . 


0.94 


Bell-metal 

Cork 


0.67 
.65 


Hard-baked clay .... 


.89 
.81 




.41 


Limestone 

Steel, hardened .... 

Cast iron 

Steel, soft 


.79 
.79 
.73 
.67 


Lead 

Clay, just malleable by the ) 
hand $ 


.20 
.17 



255. EXAMPLES. 

Ex. 1. Two inelastic bodies, weighing 12 lbs. and 7 lbs. re- 
spectively, move in the same direction with velocities of 8 feet 
and 5 feet in a second. Find the common velocity after impact ; 
also the velocity lost by one, and that gained by the other. 

Ex. 2. A mass m x , with a velocity 11, impinges onm 2 moving 
in the opposite direction with a velocity of 5 ; by impact m x 



152 DYNAMICS. 

loses one third of its momentum. What are the relative mag- 
nitudes of m x and m 2 l 

Ex. 3. m li weighing 8 lbs., impinges on m 2 , weighing 5 lbs., 
and moving in m^s direction with a velocity of 9; by impact 
m^s velocity is trebled. What was m^s velocity before im- 
pact? 

Ex. 4. Two bodies m 1 and m 2 are moving in the same 
straight line with velocities v l and v 2 . Find the velocity of 
each after impact when 6?n 1 =5m 2f v l =l i 4v 1 +5v 2 =0, and 

Ex. 5. Two bodies m x and m 2 are perfectly elastic and 
move in opposite directions; m 1 is treble of m 2 , but m 2 s 
velocity is double that of m x . Determine their motions after 
impact. 

Ex. 6. There is a row of perfectly elastic bodies in geomet- 
rical progression whose common ratio is 3 ; the first impinges 
on the second, which transmits its velocity to the third, and so 
on ; the last body moves off with ^\ the velocity of the first. 
What was the number of bodies ? 

Ex. 7. m^—Sm^ impinges on m 2 at rest; w,'s velocity 
after impact is § of its velocity before impact. Required the 
value of e, the modulus of elasticity. 

Ex. 8. Two bodies m, and m 2 , whose elasticity is f, moving 
in opposite directions with velocities 25 and 16 respectively, 
impinge directly upon each other. Find the distance between 
them 4^ seconds after impact. 

Ex. 9. At what angle must a body whose elasticity is J be 
incident on a perfectly hard plane, that the angle made by its 
path before and after impact may be a right angle ? 

Ex. 10. A ball whose elasticity is e, projected from a given 
point in the circumference of a circle, after two reflections 
from the interior of the circle, returns to the same point. Re- 
quired the angle 6 made by the direction of projection with the 
radius at the given point. 



Ans. Tan. 6- 



Vl+e+e 2 ' 



CHAPTER III. 

MOTION FROM THE ACTION OF A CONSTANT FORCE. 

256. By the definition of a constant force (Ai*t. 222), the 
velocities generated in equal successive intervals of time by 
the action of the force are all equal, and the increment of ve- 
locity in a unit of time is a measure of the force. Hence the 
velocity is the same at no two successive instants, and, if the 
body or point move from rest, will increase uniformly, or be 
uniformly accelerated. 

By the velocity acquired in t seconds is meant the space 
over which the body would pass in the second next succeed- 
ing the t seconds if the velocity should remain the same during 
this second as it was at the end of the t seconds ; or the space 
described by the body in the interval between t seconds and 
£+1 seconds, if, at the end of t seconds, the force should cease 
to act. Putting/ for the force, and v for the velocity acquired 
in the time Z, we have, as before, 

v=ft. (53) 

257. Prop. To find the space in terms of the force and time 
when a body moves from rest by the action of a constant force. 

Let/ be the force, and s the whole space described in the 

time t, and let t be divided into n equal parts, each =-. The 

n 

intervals reckoning from the commencement of the time, will 

be 

t 2t 3t 4t (n—l)t nt 

-, —, —j — , &c, , — . 

n n n n n n 

By (53), the velocities at the end of these intervals will be 

t 2t St At ,(n-l)* M 



154 DYNAMICS. 

If now the bod)'' moved uniformly during each interval - 

with the velocity it had at the beginning or end of this interval, 
the spaces described during the intervals respectively, would 
be equal to the product of this uniform velocity by the interval 
(37), and the whole space described would be equal to the sum 
of these partial spaces. 

If, therefore, the body moved uniformly during each interval 

- with the velocity it had at the beginning of the interval, we 

should have 

s=0+f.— +f—+f— +, &c, . . ./. i — •, 

J n J n J n J n 



=/■ 



n 

f n(?i-l) 

n*' 2 ' 



=f-2-f'2n' <"> 

But if the body moved uniformly during each interval - 

with the velocity it had at the end of the interval, we should 
have 

= /v[s(l+2 + 3+, &c, ....,-«). 
= f n(n+l) 

=/|+/£. © 

Since the velocity is uniform during no sensible interval, the 
true value of s will lie between the two quantities (a) and (b), 
however small each interval may be, or however large n may 
be. But when n becomes indefinitely large, the last terms in 
(a) and (b) vanish, and (a) = (b). 



ACTION OF A CONSTANT FORCE. 155 

.-. s=±ff and sarf. (54) 

Hence the space described from rest by a body from the 
action of a constant force is equal to half the product of the force 
by the square of the number of seconds, and the spaces vary as 
the squares of the times. 

258. Prop. To determine the space in terms of the force and 
velocity ; also in terms of the time and velocity. 

1°. Eliminating t from (53) and (54), we have 

s=-^-f- or socv 2 . (55) 

2°. Eliminating/ from (53) and (54), we obtain 

s=^vt, or sccvt. (56) 

Cor. The space described in any time by a body moving 
from rest by the action of a constant force, is half that it would 
describe in the same time if it moved uniformly with the ac- 
quired velocity. 

For the space s, described in the time t, with a uniform ve- 
locity v, is, by (37), 

s 1 =vt, 
which, compared with (56), gives 

259. Prop. To find the space described by a body in the last 
n seconds of its motion. 

The space described in the time t (54) is 

s,=iff. (a) 

The space described in the time (t—n) is 

Subtracting (b) from (a), we have, for the space s described 
in the last n seconds, 

s= 5l -s 2 =l/(2^-7i 2 ). (57) 

Cor. If the space described in the m seconds next preceding 
the last n seconds is required, we have, for the space s 3 , de- 
scribed in the time t— (n+m) seconds, s 3 =^f(t—n—m)\ which, 
subtracted from (6), gives for the space s required, 

s=s 2 -s z =if(2mt-2mn-m 2 ). (58) 



156 DYNAMICS. 

260. Prop. A body being projected with a given velocity in 
the direction in which a constant force acts, to find the velocity 
of the body at the end of a given time, and the space described in 
that time. 

By Art. 234, the velocity due to the joint action of the im- 
pulsive and constant forces will be equal to the sum or differ- 
ence of the velocities due to each, according as they act in the 
same or opposite directions. If, therefore, v 1 be the velocity 
of projection, the whole velocity v, at the end of the time t, 
will be 

v=v,±ft. (59) 

In the same manner, the space due to the joint action of the 
forces will be equal to the sum or difference of the spaces due 
to each, or 

s=v x t±\ff. (60) 

261. Prop. A body being projected with a given velocity in 
the direction in which a constant force acts, to find its velocity 
when it has passed through a given space. 

Let s be the given space, and h the space through which the 
body must pass to acquire the velocity v l by the action of the 
constant force. Then (55) 

vl=2fh, 
and for the space (h±s) 

v 2 =2f(h±:s), 
=2fh±2fs, 
=v\±2fs, (61) 

the signs to be taken as in the last proposition. 

262. Prop. When a body is projected in a direction opposite 
to that in which a constant force acts, the velocity acquired in 
returning to the point of departure is equal to the velocity of 
projection. 

If v j be the velocity of projection in the direction AB, from 
(61) we have 

5= -2T {a) 



ACTION OF A CONSTANT FORCE. 157 

Now the actual velocity v of the body is continual- 
ly diminished by the action of the force/, and when 
the body has reached its greatest distance from A, as 
B, v=0. Hence (a) gives 

2 

s=^7, or v,= V 2fs. 

But the velocity v acquired in moving from rest 
through BA=s is, by (55), 

v= V2fe. , 

Cor. 1. The velocity of the body at any given dis- 
tance from A is the same in going and returning. 

For in the expression (a), since v l and 2f are constant, v is 
always the same for the same value of s=AC 

If v>v lf s is negative, or on the opposite side of A from B. 

2v 
Cor. 2. The whole time of flight T=-r 1 . Making v=0 in 

v 
(59), we have t=-ji-, when the body is at B. 

But, in returning, it acquires the velocity v 1 =ft. 

v 
Hence the time of return is t=-7?, 

2v, 
and the whole time 2t=T=—^-. 

263. Schol. 1. There is no known instance in nature of a force 
which is constant. The law of Universal Gravitation is, that 
every particle of matter attracts every other particle with a 
force which varies directly as the mass of the attracting particle, 
and inversely as the square of the distance. A sphere of uni- 
form density, or one whose density is the same at equal distan- 
ces from the center, attracts a body exterior to it as if the mat- 
ter of the sphere were collected at its center, and with a force 
varying inversely as the square of the distance of the body from 
the center, but a body or particle in the interior with a force 
varying directly as its distance from the center. 

Regarding the earth as a sphere, this is the law of the earth's 



158 DYNAMICS. 

attraction for bodies exterior to it ; but for all small distances 
above the surface the intensity of gravity may be considered 
constant, since at a distance of one mile above the surface the 
actual diminution of gravity is only 197 j. 3Ui , or about the 2000th 
part of that at the surface ; a variation too small to affect sens- 
ibly the circumstances of the motion of a falling body com- 
puted on the hypothesis that the force suffers no variation at 
all. 

In reality, the force by which a body is drawn toward the 
earth is equal to the sum of the attractions of each for the 
other ; but when the mass of the body is inconsiderable with 
regard to the mass of the earth, the effect of the former is in- 
sensible, and the accelerations of all bodies of moderate size 
are the same. Within these limits, then, gravity may be taken 
as a constant or uniformly accelerating force, whatever be the 
mass. 

264. Schol. 2. Representing the intensity of gravity at the 
surface of the earth, as before, by g, we have, from (54), 
s=lgf, and if t=l, s=^g, or g—2s; from which it appears 
that the acceleration is equal to twice the space described in 
the unit of time. It has been found by experiment that the 
space through which a body falls freely in one second in the 
latitude of New York is equal 16.0799 feet, or 16^ feet nearly. 
Hence g-=32.1598 feet, or 32£ feet nearly. Hence, also, all 
the relations between the space, time, and velocity due to the 
action of a constant force are true in relation to the action of 
gravity near the earth's surface. Collecting these, and substi- 
tuting g for /, we have the following relations, supposing no 
resistance from the air : 

s=hgt*=^,=¥v, (54') 

2$ 

v 2s /2s 
t= - =—=*/—, (56') 

g v v g v ' 

v 2s y 2 
8= t =F=2? < 55 > 



ACTION OF A CONSTANT FORCE. 159 

Also, when a body is projected with a velocity v x vertically 
upward or downward, (260) and (261), 

s=v,t±\gt% (60') 

v'=vl ±2gs. (6F) 

265. EXAMPLES. 

Ex. 1. A body has been falling 11 seconds. Find the space 
described and the veloctiy acquired. 

By (540, 5=%£ 2 =16 T VX121 = 1946 T V f ee t. 
By (53'), v= gt =32} X 11= 353f feet. 

Ex. 2. Find the time in which a falling body would acquire 
a velocity of 500 feet, and the height from which it must fall. 

v 500 
By (560, t= - =^-7 = 15.544 seconds. 
J V g 32j 

v 2 250000 

By (540, s =^=-^T =S8m feetnearl y> 

or s =itv=±x 15.544X500=3886 feet, as before. 

Ex. 3. What is the velocity acquired by a body in falling 
450 feet ? and if the body weigh 10 tons, what is the momentum 
acquired ? 

Ans. v =170.14 feet. 

M=3811136 lbs. 

Ex. 4. A body had fallen through a height equal to one 
quarter of a mile. What was the space described by it in the 
last second? 

Ans. 5=275 feet. 

Ex. 5. A body had been falling 15 seconds. Compare the 
spaces described in the seventh and last seconds. 

Ex. 6. A body had been falling 12.5 seconds. What was 
the space described in the last second but 5 of its fall? 

Ex. 7. The space described by a body in the fifth second 
of its fall was to the space described in the last second but 4, 
as 1 to 6. What was the whole space described? 

Ans. s= 15958.69 feet. 

Ex. 8. A body is projected vertically downward with a ve- 



160 DYNAMICS. 

locity of 100 feet. What is its velocity at the end of 5 sec- 
onds? 

Ex. 9. A body is projected vertically upward with a veloci- 
ty of 100 feet. Find its velocity at the end of 5 seconds, and 
its position at the end of 8 seconds. 

Arts. v= — 60| feet. 
s=-229% feet. 

Ex. 10. A body is dropped into a well and is heard to strike 
the bottom in 4 seconds. What is the depth of the well, the 
velocity of sound being 1130 feet? 

Ans. 231 feet. 

Ex. 11. A body is thrown vertically upward with a veloci- 
ty Dj. Find the time at which it is at a given height h in its 
ascent. 

If t be the time required, (60) gives 

■ %v x 2h 
whence t— t — H =0, 

g 8 



v,± Vv 2 .—2gh 

or t =— - 2-. 

g 
The lower sign gives the time when the body is at the 
height h in its ascent, and the upper in its descent. 

Ex. 12. A body is projected vertically upward, and the in- 
terval between the times of its passing a point whose height is 
h in its ascent and descent is 2t. Find the velocity v of pro- 
jection, and the whole time T of its motion. 

Ex. 13. A body whose elasticity is e is projected vertically 
upward to the height h above a hard plane, to which it returns, 
and from which it rebounds till its motion is destroyed. What 
is the whole space described by the body ? 



CHAPTER IV. 



PROJECTILES. 

In the preceding chapter we have discussed the motion of 
a body by the joint action of a projectile force and the force 
of gravity, when these forces were coincident or opposite in 
direction. We now proceed to determine the circumstances 
of the motion of a body, when the direction of the projectile 
force is other than vertically upward or downward, supposing, 
as heretofore, no resistance from the air. 

266. Prop. The path of a body, moving under the joint in- 
fiuence of a projectile force, and the force of gravity considered 
as a constant, accelerating force, is a parabola. 

Let v be the velocity of pro- 
jection from the point A, in the 
direction ANY, and t the time 
in which the body will describe 
AN=y, with the uniform veloci- 
ty v, if gravity do not act. Let 
AM.=x be the space through 
which the body will fall in the 
time t by the action of gravity. 
Completing the parallelogram 
MN, the actual place of the 
body at the end of the time tis ^ 
P. 

By (37), &N=y=vt; and by 
(54),AM=NP=a J =fe* a . Elim- X 
mating t from these two equa- 
tions, we get 

y— — x. (a) 

If h be the space due to the velocity v, or the space through 

L 




162 



DYNAMICS. 



which the body must fall to acquire the velocity of projection 
(55), v 2 =2gh, which, substituted in (a), gives 

y*=4hx, (62) 

the equation of a parabola referred to the oblique axes AX, 
AY ; and, since 4h is the parameter to the diameter through 
A, h is the distance from A, the point of projection to the focus 
F, or to the directrix. 

Cor. The velocity of the body at any point of its path is 
that which the body would acquire in falling vertically from 
the directrix to that point. 

For if the body were projected from any point of its path in 
the direction, and with the velocity it has at that point, it would 
obviously describe the same path. Therefore the velocity at 
that point must be equal to that due to one fourth the parame- 
ter to the diameter at that point, which is the distance from 
that point to the directrix or to the focus. 

267. Prop. To find the equation to the path of a projectile 
when referred to horizontal and vertical co-ordinate axes. 

Let v be the velocity of pro- 
jection in the direction AN, 
which makes with AX the 
angle of elevation NAX=<z, 
APB the path of the body, 
AM=i, PM=y, the co-or- 
dinates of the point P, t the 
time in which the body de- 
scribes the arc AP, and pro- 
duce MP to meet the direc- 
tion of projection in N. 
Then AN =vt, NP=%f , and MN=vt sin. a. 
Also, AM=x=vt. cos. «, («) 

and PM =y=vt sin. a-\gt\ (b) 

Eliminating t from (a) and (b), we obtain 

- y—x tan. a— x 2 -— ^ =— , (63) 

* 2v 2 cos.« v 

and, substituting the value ofv 2 =2gh, 



Y 


/ 

5^" "^ 




L \ 




\x 



A 



M 



PROJECTILES. 163 

268. Def. The horizontal range of a projectile is the distance 
AB from the point of projection to the point where it strikes 
the horizontal plane in its descent. The time of flight is the 
time occupied in describing APB. The height through which 
the body must fall to acquire the velocity of projection is call- 
ed the impetus. 

269. Prop. To find the time of flight of a projectile on a hori- 
zontal plane. 

At the points A and B the ordinate y=0. This value of y, 

substituted in (b), Art. 267, gives 

vt sin. a— ±gf=0. 

2v sin. a , ■ 

.'. t=0, and t= . (65) 

o 

The former value of t applies to the point A, and the latter 

to the point B, which is, therefore, the time of flight required. 

Or, since v sin. a is the vertical component of the velocity, or 

the velocity of projection estimated vertically, if this value of 

the velocity be substituted for v 1 in (60), and we make s=0, 

we get, as before, 

0=vt sin. a— \gf, 

2v sin. a 
t =0, and t= . 

g 

270. Prop. To find the range of a projectile on a horizontal 
plane. 

Put y=0 in (64), and we have 

<r 2 

0=x tan. a- 



4h cos. 3 a 

from which we obtain 

for the point A, x=0, 

for the point B, x =4h sin. a cos. a, 

or AB=R=2A sin. 2a, (66) 

the horizontal range required. 

Cor. 1. The horizontal range is greatest when a=45°. 
For in this case 2a=90°, and sin, 2a=l. .*. x=2h. 




1G4 DYNAMICS. 

or the greatest horizontal range is equal to twice the height 
due to the velocity of projection, or twice the distance from 
the point of projection to the focus of the trajectory. 

Cor. 2. The range is the same for any two angles of eleva- 
tion, the difference between which and 45° is the same, or for 
(45±0). 

For sin. (90° +20) = sin. (90°-2(9), 

or sin. 2(45° +$) = sin. 2(45°-0). 

If, therefore, we put either 45° +6 or 45°— 6 for a in (66), 

the value of R remains the 
same. Thus, if AB be the 
range of a projectile when 
the angle of elevation is 45°, 
and APB its path, the range 
AB' will be that due to the 
elevation 45° — d, for which 
B ' B the path is APB', and to 

45°+0, for which the path is AP"B'. 

Cor. 3. When the velocity of projection is given, and we 
know the range R due to the elevation a, we can readily find 
the range R' due to any other elevation a'. 

For R=2/isin. 2a, 

and R f =2hsia. 2a'. 

... R<=£^.R. (67) 

sin. 2a v ' 

Cor. 4. Since the horizontal velocity of a projectile is uni- 
form, the range is equal to the horizontal component of the ve- 
locity into the time of flight, or 

„ 2v sin. a 2v 2 sin. a. cos. a. _ . - . . 

Yt=v. cos. aX -= —=2h sin. 2a, as before. 

g S 

271. Prop. To find the greatest height which a projectile at- 
tains. 

The greatest height H is evidently the value of the ordinate 

at the middle of AB, or when the time is one half the time of 

_ v sin. a . ... . 
flight, Putting t—- m (b), Art. 267, we have 

o 



PROJECTILES. 



165 



D'sm. a jV sin. a 
H= 2 i : 



1,,» 



v sin. a 



=£ sin. 2 a. 



(68) 



272. Prop. To find the co-ordinates of the point where a pro- 
jectile will strike an inclined plane passing through the point 
of projection, the range on the inclined plane, and the time of 
flight. 

Let y—x tan. (3 be the equation 
of the line AC, which is the in- 
tersection of the inclined plane, 
with the vertical plane of the 
path of the body. 

Substituting this value of y in 
(64), we obtain, after reduction, 
for the abscissa of the point C, 

_4/i cos. a. sm.(a—0) 




cos. (3 

By substituting in (64) for x, its value y cot. (3, and reducing 
we get for the ordinate of C, 

_4h. cos. a. sin. (3. sin. (a—f3) 
cos. 2 ]3 

1 



y- 



To find AC=R', multiply (a) by sec. (3= 



cos. 0' 



(b) 

and we 



have 



A ^ _ f _ Ah. cos. a. sin. (a— 0) , x 
AC=R , =a; sec. 0= ^ ^. (69) 

cos. 2 |3 v ' 



If the inclined plane cut the path of the projectile below the 
axis AX, (3 will be negative. 

The time of flight is equal to the time of describing the ab- 
scissa AM with the horizontal component of the velocity. 
Hence (37), dividing (a) by v cos. a., we get 

4/isin. (a— (3) 
v. cos. j3 ' 



2v sin. (a— (3) 
g. cos. (3 



(70) 



166 DYNAMICS. 

273. Schol. It has been found by experiment, that if w be 
the weight of a ball or shell, p the weight of the gunpowder 
used in discharging the ball or shell from a mortar, and v the 
velocity generated by the powder, 



t>=1600y- 



feet. (71) 

w 



274. EXAMPLES. 

Ex. 1. A body is projected at an angle of elevation of 15° 
with a velocity of 60 feet. Find the horizontal range, the 
greatest altitude, and time of flight. 

, x , v 2 3600 
From (54'), h== ^=-^r= 55 ' 96 ' 

From (66), R=2h s'm.2a=2h. sin. 30° =2.h.\=h= 55.96 
From (68), H= h sin. 2 a. Log. sin. a= 9.4129962 

" " = 9.4129962 

h = 1.7478777 

.*. H=8.7486 " H=.0.5738701 



_ , v m 2v sin. a 

From (65), T= -. Log. 2v= 2.0791812 

o 

Log. sin. a= 9.4129962 

ax. " g= 8.4925940 

.-. T=0.96554 " T= 9.9847714 

Ex. 2. A body is projected at an angle of elevation of 45°, 
and descends to the horizon at a distance of 500 feet from the 
point of projection. Required the velocity of projection, the 
greatest altitude, and time of flight. 

Ans. v =126.82 feet. 
H=125. 

T=5.58 seconds. 
Ex. 3. The horizontal range of a projectile is 1000 feet and 
the time of flight is 15 seconds. Required the angle of eleva- 
tion, velocity of projection, and greatest altitude. 

Ans. a =74°.33 / .09". 
v =250.29 feet. 
H=904.69 " 



PROJECTILES. 167 

Ex. 4. If a body be projected at an angle of elevation of 
60°, with a velocity of 850 feet, find the parameter to the 
axis of the parabola described, and the co-ordinates of the 
focus. 

Ans. p= 11230.57 feet. 
x= 9725.67 " 
y= 5615.28 " 

Ex. 5. Find the velocity and angle of elevation of a ball that 
it may be 100 feet above the ground at the distance of one 
quarter of a mile, and may strike the ground at the distance 
of one mile. 

Ans. a=5°.46'.04 /, .6. 
^=921.566 feet. 

Ex. 6. What must be the angle of elevation of a body in 
oraer that the horizontal range may be equal to three times the 
greatest altitude ? What, that the range may be equal to the 
altitude ? 

Ex. 7. A body is projected at an angle of elevation of 60°, 
with a velocity of 150 feet. Find the co-ordinates of its posi- 
tion, its direction, and velocity at the end of 5 seconds. 

Ex. 8. A body is projected from the top of a tower 200 feet 
high, at an angle of elevation of 60°, with a velocity of 50 feet. 
Find the range on the horizontal plane passing through the 
foot of the tower, and the time of flight. 

Ex. 9. A body, projected in a direction making an angle of 
30° with a plane whose inclination to the horizon is 45°, fell 
upon the plane at the distance of 250 feet from the point of pro- 
jection, which is also in the inclined plane. Required the ve- 
locity of projection, and the time of flight. 

Ex. 10. The heights of the ridge and eaves of a house are 
40 feet and 32 feet respectively, and the roof is inclined at 30° 
to the horizon. Find where a sphere rolling down the roof 
from the ridge will strike the ground, and also the time of de- 
scent from the eaves. 

Ex. 11. How much powder will throw an eight-inch shell, 
weighing 48 lbs., 1500 yards on an inclined plane, the angle of 



168 DYNAMICS. 

elevation of the plane being 28°.45', and that of the mortar be- 
ing 48°.30 / ? 

Ex. 12. Find the velocity and angle of elevation that a pro 
jectile may pass through two points whose co-ordinates are 
z=300 feet, y=60 feet, £'=400 feet, and y'=40 feet. Also 
find the horizontal range, greatest altitude, and time of flight. 



CHAPTER V. 




CONSTRAINED MOTION. 
§ I. MOTION ON INCLINED PLANES. 

275. Prop. To determine the relations of the space, time, and 
velocity when a body descends by the action of gravity down an 
inclined plane. 

Let the body fall from C down C 

the inclined plane CA, whose in- M 

clination is a, M be the position of 
the body at any time t from rest at 
C, CM=s, and v= the velocity at A- 
M. 

The force of gravity g, acting in the direction Mg, may be 
resolved into two forces, one f=g sin. a acting in the direction 
CA, the other g cos. a acting perpendicularly to C A, and wholly 
ineffective in producing motion ; the body is, therefore, urged 
down the inclined plane by the constant accelerating force 
f=g. sin. a. 

If, therefore, this value of/ be substituted 
in (53), we have v =gt sin. a, (72) 

in (54), we have s =\gt* sin. a, (73) 

in (55), we have v*=2gs sin. a, (74) 

from which all the circumstances of the motion may be de- 
termined. 

276. Prop. The velocity acquired by a body in falling down 
an inclined plane is equal to that acquired in falling freely 
through the height of the plane. 

If s=AC, the length of the plane, and A=BC, the height, by 
(74), 

v 2 =2g.s. sin. a 
=2g.CB. 



170 



DYNAMICS. 



.*. V = V2gh, 

which, by (53'), is the velocity due to h, the height of the 
plane. 

277. Prop. The times of descent down different inclined planes 
of the same height vary as the lengths of the planes. 

By (73), s=igt 2 sin. a, 

BC 

or AC=igl'j£. 

••• < =Ac \4fc 

<r AC, when BC is constant. 

278. Prop. To find the relations of the space, time, and veloc- 
ity when a body is projected down or up an inclined plane. 

Substituting for /, its value, g sin. a, on an inclined plane, 
in (59), we have v =v x ±gt sin. a, (75) 

in (60), we have s =v x t±^gf sin. a, (76) 

in (61), we have v 2 =v\±2gs sin. a, (77) 

which give all the circumstances of the motion of the body. 

279. Prop. The times of descent down all the chords of a cir- 
cle in a vertical plane, drawn from either extremity of a vertical 
diameter, are the same, and equal to that down the vertical di- 
ameter. 

Let AB be the vertical diameter, BD 
and AD any chords drawn from its ex- 
tremities, and DC perpendicular to AB. 
To find the time of descent down BD, 
we have, from (73), 

BD=^ 8 sin. BDC 
BC 
BD* 
2BD 2 
JSC 
2.BA 
g 




_i 



f=- 



CONSTRAINED MOTION. 



171 



or 



/2.BA 



which, by (56'), equals the time down the diameter. 
In the same manner, 

CA 

from which we obtain, as before, 



-4 



2.BA 




280. Prop. To find the straight line of quickest descent from 
a given point within a vertical circle to its circumference. 

Let P be the given point. Draw the 
vertical radius CA, join AP and produce 
it to meet the circumference at B ; PB 
will be the line required. 

Join BC and draw POD parallel to 
AC. Since BC=AC, BO=PO. With 
O as a center, and radius PO, describe 
the circle PBD, which will be tangent to 
the circle C at B. 

Now the time down PB will equal the time down the verti- 
cal diameter PD, and the time down every other chord drawn 
from P {Art. 279). But any other chord from P, produced to 
the circumference of C, will be partly without the circum- 
ference of O, and, therefore, the time down it will be greater 
than the time down PB, which is therefore the line of quickest 
descent to the circumference of C. 

281. Prop. To find the straight line of quickest descent from 
a given point to a given inclined plane. 

Let A be the given point, AG a vertical 
line passing through this point, and AB a 
perpendicular to the inclined plane BG. 
The line AC, bisecting the angle BAG, 
will be the line required. 

Draw CE parallel to BA, and therefore 
perpendicular to BG. Since EAC=CAB 




172 DYNAMICS. 

=ECA, the triangle EAC is isosceles. With E as a center, 
and EA or EC as a radius, describe the circle ACD, to which 
BCG will be a tangent. 

Now the time down AC will be equal to the time down AD, 
and to the time down every other chord drawn from A. But 
any other chord of the circle must be produced to meet the 
plane BG. Therefore the time down any other line, drawn 
from A to the plane, will be greater than the time down AC, 
which must be the line of quickest descent required. 

282. Prop. Two bodies are suspended from the extremities of 
a cord passing oner a fixed pulley : to determine the circumstances 
of their motion. 

» Let P and Q be the weights of the bodies, of which 

o jj P is the greater. By (22) their masses are — and — 

respectively. Neglecting the rigidity of the cord, 
the inertia and friction of the pulley, if P=Q, they 
will counterpoise, and no motion will ensue ; but 
P>Q, P will descend, and Q, rise through equal 
spaces by a force equal to the difference P— Q, of 
their weights. But this being a moving force (Art. 
225), is equal to the accelerating force into the mass 

moved, =/( ), where/ represents the accelerating 

force. 

Hence p_Q=/£±^, 

5 

This being a constant accelerating force, by substituting its 
value in (53), (54), and (55), we have expressions for the 
space, time, and velocity. 

283. Schol. If the inertia I of the pulley be taken into con- 
sideration, Ig is an additional force to be overcome by the 
difference of the weights, and in this case 



© 



® 



CONSTRAINED MOTION. 173 

This is the formula for Atwood's machine, an instrument for 
illustrating the laws of falling bodies. The friction of the pul- 
ley is reduced by friction wheels, and the rigidity of the cord 
by employing a fine flexible thread. By making the difference 
between P and Q small, the motion is made so slow as to ren- 
der the time, space, and velocity easily determinable. 

§ II. MOTION IN CIRCULAR ARCS. 

284. Prop. When a body descends by the action of gravity 
down any smooth arc of a circle in a vertical plane, the velocity 
at the lowest point is proportional to the length of the chord of 
the arc. 

Let a body descend from the point D (Fig., Art. 279) down 
the arc DA. Since the reaction of the curve is always per- 
pendicular to the path of the body, it can neither accelerate 
nor retard the motion of the body. Its velocity, therefore, at 
A will be that which it would acquire in falling through the 
vertical height C A of the arc. 

Hence, bv (55), v*=2gCA. 

AD 2 

=2g- 



.-. v=AB^J 2 ' 



AB 



<T AD. 

When the arcs are small, the velocities are nearly proportional 
to the arcs, the principle referred to in Art. 254. 

285. Prop. When a body is constrained to describe the sides 
of a polygon successively, to find the velocity lost in passing from 
one side to the succeeding one. 

Let AB and BD be two adjacent sides 
of the polygon, w the angle made by their ^- 
directions, and v the velocity of the body 
at the point B. In passing from AB to E> 

BD some of the velocity will necessarily be lost. Resolving 
the velocity in the direction BD, we have, for the velocity in 
BD, 




174 DYNAMICS. 

V COS. 10. 

This, subtracted from the primitive velocity v, will give for the 
velocity lost, 

v 1 =v— v cos. w 

—v(l — cos.w) 

—v versin. w 



2— versin. w 

286. Cor. If the sides of the polygon be increased in num- 
ber, the angle w diminishes ; and when the polygon becomes a 
circle, w, and consequently its sine, becomes indefinitely small. 
In this case, versin. w, small in comparison with 2, may be re- 
jected, and 

v • a 
v 1 =- sin. w. 

But if sin. w is infinitely small, sin. 3 w is infinitely smaller, and 
the velocity lost 

When, therefore, a point or body is constrained to describe a 
curve, no velocity is lost by the reaction of the curve. 

287. Prop. If a material point move through one side of a 
regular polygon with a uniform velocity, to find the direction 
and intensity of the impulse which must be given to the material 
point at each angle of the polygon in order that it may describe 
the entire polygon with the same uniform velocity. 

Let ABC ... be the polygon, and let 
the material point describe AB with the 
velocity v in the unit of time. If, when 
at B, no other force act upon the point, 
it will describe BD=AB in the same 
time. Join DC, and draw BN equal 
and parallel to DC. If, when at B, the 
point receive an impulse in the direc- 
tion BN, such as to cause it to describe 
BN in the same time that it would describe BD, the point will 
describe the diagonal BC, the succeeding side of the polygon, 




CONSTRAINED MOTION. 175 

in the same time. But since BC=BD, the velocity in BC is the 
same as that in AB, and the point will describe the two adjacent 
sides with the same uniform velocity. Now, since the trian- 
gle BCD is isosceles, BDC=BCD=CBN. But ABN=BDC ; 
.♦. ABN=CBN, and BN bisects the angle ABC. Hence the 
direction of the impulse will pass through the center of the cir- 
cumscribing circle, and, if similar impulses be applied at each 
of the angles of the polygon, the point will describe the entire 
polygon, with the original velocity unchanged. 

To find the magnitude of the component force BN==f, let r 
be the radius of the circle ; and, since BN is perpendicular to 
the chord AC, 

BC 2 

and f=BN= . 

: r 

But BC=AB is the space described in the unit of time, and is 
therefore represented by v. Hence 

or, the intensity of the impulse is equal to the square of the ve- 
locity in the polygon divided by the radius of the circumscribing 
circle. 

288. Cor. The force which must continually urge a material 
point toward the center of a circle, in order that it may de- 
scribe the circumference with a uniform velocity, is equal to 
the square of the velocity divided by the radius. 

Since the reasoning in the proposition is independent of the 
number of sides in the polygon, the sides of the polygon may 
be increased in number, and the frequency of the impulse in- 
creased in the same ratio, without affecting the relation of the 
intensity of the impulse to the velocity in the polygon. But 
when the number of sides becomes infinite, or the polygon be- 
comes a circle, the impulses will no longer be successive, but 
an incessant action of the same intensity. 

The force is, therefore, a constant accelerating force, and 



176 DYNAMICS. 

If the mass of the body be taken into consideration, the 
moving force {Art. 225) will be 

F-/=^. (80) 

It is obviously immaterial whether the body be retained in 
its path by the reaction of a smooth curve, or by an inextens- 
ible cord without weight, by which it is connected with the 
center of the circle. F will be a measure of the resistance of 
the curve in the one case, and of the tension of the cord in the 
other. 

289. Def. The force which constantly urges a body toward 
the center of its circular path is called a centripetal force. The 
tendency which the body has to recede from the center, in con- 
sequence of its inertia, or the resistance which it offers to a de- 
flection from a rectilinear path, the resistance being estimated 
in the direction of the radius, is called a centrifugal force. 

290. Prop. To discuss the circumstances of the motion of a 
body constrained to move in a circle by the action of a central 
force. 

1°. When the masses are equal, by (79), 

r 

If r be constant, 

fccv 2 ; 

or, when a body moves in the circumference of a circle by 
means of a cord fixed at the center, the tension of the cord, or 
the centrifugal force, will vary as the square of the velocity. 
2°. If v be constant, 

J r 

or, if equal bodies describe different circles with the same ve- 
locity, the centrifugal forces will be inversely as the radii of the 
circles. 



CONSTRAINED MOTION. 177 

3°. Let T be the periodic time, or time of one revolution. 

Then (37) vT=2tt^ 9 

' a 4ttV 2 

or v =-7p-. 

(79), =fr. 

' 4?rV 
•'• /="tT. (81) 

r 

or, the centrifugal force varies directly as the radius of the circle, 
and inversely as the square of the periodic time. 
4°. If T oc r\ by substitution in (81), 
, r 1 

or, when the squares of the periodic times are as the cubes of 
the distances from the center, the centrifugal force will be in- 
versely as the square of the distance. 

5°. If w be the angle subtended by the arc described by the 
body in the unit of time, o) is called the angular velocity. As- 
suming for the angular unit the angle whose arc is equal in 
length to the radius, v=ro>. Substituting this value of v in 

(79). 

f=rtt; (82) 

or, the centrifugal force varies as the product of the radius and 
the square of the angular velocity. 

6°. If in each of the above cases the masses be not the same, 
the centrifugal forces will vary directly as the masses in addi- 
tion to the other causes of variation. 

The foregoing principles admit of a very simple and satis- 
factory illustration by means of an instrument called the whirl- 
ing table. 

291. Prop. To find the relation of the centrifugal force in a 
circle to the force of gravity. 

Let h be the height due to the velocity v which the body 
has in the circle. By (53'), v'=2gh. This value of v\ substi- 
tuted in (79), gives 

M 



178 DYNAMICS. 

/- r > 

f 2h 
or J -= — . 

g r 

Hence the centrifugal force is to the force of gravity as twice 
the height due to the velocity in the circle is to the radius of the 
circle. 

Cor. If f=g, r=2h, 

or, in a circle whose radius is equal to twice the height due to 
the velocity in the circle, the centrifugal force is equal to the 
force of gravity. 

292. Prop. To find the centrifugal force at the equator. 

By (8i), f=^. 

The equatorial radius R of the earth is 3962.6 miles =20,922,528 
feet. tt=3.1415926. And, since the earth revolves on its axis 
in 0.997269 of a day, T = 0.997269X86400 8 =86164 3 . These 
values, substituted in the above, give 

/=0.1 11255 feet. 

293. Schol. Since the force of gravity g at the equator has 
been found to be 32.08954 feet, if G be the force of gravity on 
the supposition that the earth does not revolve on its axis, then 

g=G~f, 
or G=g +/; 

= 32.08954+0.111255=32.200795, 

/ 0.111255 1 

and G= 32.200795 =289 nearly; ^ 

or the centrifugal force at the equator is — — the force of 

<^89 

gravity. 

294. Prop. To find the time in which the earth must revolve 
on its axis, in order that the centrifugal force at the equator 
may equal the force of gravity. 

Let T' be the required time, and/' the corresponding cen- 
trifugal force. From (79) we have 



CONSTRAINED MOTION. 



179 



II R^ 

np2 * rp/2' 

But in this case R=R', and/' is to become equal to G. 

T12 * r T'/2» 



/=/' : 



f:G=-, 



or 



and 



f 1 

rp/2 ^Lnna 'T's 

1 ~G "289 1 ' 
T' =— T nearly. 



Hence the earth must revolve in — th of its present period. 

295. Prop. The centrifugal force diminishes gravity at dif- 
ferent places on the earth's surface in the ratio of the square of 
the cosine of the latitude. 

Regarding the earth as a sphere, from 
which it differs but very little, let EF be 
the axis, AC=R the equatorial radius, B 
any point whose latitude is the arc AB, 
measured by the angle ACB=</>, and 
BG=R' the radius of the parallel of lati- 
tude passing through B. Let the centrif- 
ugal force at B, which acts in the direc- 
tion GB, be represented by BD. 

47T 2 R' 

By (81), BD=^. 

Resolving this in the direction of the vertical CZ, opposite to 
that in which gravity acts, and calling this component/', we 
have 

/'=B^ R 




cos. 



But R'=R cos. 0, 



••• /' : 



4tt 2 R 



COS. 2 0, 



(84) 



in which the coefficient of cos. 2 <£ is constant for all latitudes. 
Hence /' x cos. 2 <f>, 



180 DYNAMICS. 

/' being the diminution of gravity by the action of the centrif- 
ugal force. 

The latitude of Middletown being 41°.33 / .10 // , the centrif- 
ugal force/', in a vertical direction at this place, will be found 
to be 

/':=0.062305. 
If this be added to the observed gravity g*'=32.l6208, we shall 
have for the whole gravity, undiminished by the centrifugal 
force, 

G'=g-'+/ v = 32.224385. 

296. Cor. Resolving the centrifugal force at B in a direc- 
tion perpendicular to the radius CB, we have 

4.7T 2 T? 

f =Bd=-y - cos. sin. <£, 

2tt 2 R . 

=-7j^- sin. 2(p. 

If, therefore, the matter of the earth were susceptible of 
yielding to this component of the centrifugal force, it would 
necessarily cause the earth to deviate from a spherical form. 
The fluid portions of the surface are therefore urged toward 
the equatorial regions, thereby increasing the equatorial diame- 
ter and diminishing the polar. If the solid portion did not also 
partake of the same general form as the fluid, we should ex- 
pect to find vast equatorial oceans and polar continents, with 
polar mountains far exceeding the equatorial in height. But 
the actual distribution of the waters and mountains on the sur- 
face of the earth is widely different. The amount of deviation 
from a spherical form, which the earth must take from the ac- 
tion of the forces developed by its motion, will depend in part 
on the law of variation of its density from the surface to the 
center. By measurement the equatorial diameter is found to 
exceed the polar about 26 miles. The ratio of the difference 
of the equatorial and polar radii to the equatorial radius, called 

the compression or ellipticity of the earth, is about — — . 

oOO 

297. Prop. To find the centrifugal force of the moon in its 
orbit. 



CONSTRAINED MOTION. 181 

Let R be the radius of the earth, nR the mean radius of the 
moon's orbit, and P the periodic time of the moon in seconds. 
Then the velocit j of the moon will be 

2nnR 
" = — ' 
This, substituted in (79), gives, for the centrifugal force, 

47T a 7lR 

If the earth were an exact homogeneous sphere, the intensity 
of gravity would be the same at all points on the surface. Be- 
ing a spheroid, differing little from a sphere, the theory of at- 
traction of spheroids shows that it is necessary to use the in- 
tensity of gravity at a latitude d>, of which the sine is the V^; 
whence 0=35°. 15'. 52". Supposing the compression of the 
earth to be 0.00324, in the latitude <j>, 

R=20897947 feet. 

To find n, conceive two lines drawn from the center of the 

moon, one to the center of the earth, and the other tangent to 

the earth at the equator. The angle made by these lines at 

the center of the moon, and subtended by the equatorial radius 

of the earth, called the moon's equatorial horizontal parallax, 

is found by astronomical observations to be at a mean 57'. 1". 

This angle, when subtended by the radius of the earth in the 

R 1 

latitude of 35°.15 , .52 // , is 7r=56'.57".33. Now -g-= sin. n=- 

nR n 

from which we find n=60.3612. P=2360585 8 . These values, 

substituted in (a), give 

/=0.00894. 

298. Cor. The moon is retained in its orbit by the gravity 

of the earth. For, since the intensity of gravity is inversely as 

the square of the distance from the earth's center, its intensity 

g 1 at the moon may be found by the proportion 

g:g '~W : ?i 2 R 2; 

p* 
whence g> =£-. 

° n 

The intensity of gravity in the latitude of 35°.15'.52", un- 



182 DYNAMICS. 

diminished by the centrifugal force of the earth in its diurnal 
motion, is 

#=32.24538. 
Substituting for n and g their values, we find 
£■'=0.00885. 

The difference between/ and g' is less than one ten thousandth 
of a foot, a difference which may be attributed to errors of ob- 
servation. 

§ III. PENDULUM. 

299. Def. A simple pendulum is a material point suspended 
by a right line, void of weight, and oscillating about a fixed 
point by the force of gravity. The path of the point is the arc 
of a vertical circle, of which the fixed extremity of the line is 
the center. 

300. Prop. To find the force by which the pendulum is urged 
in the direction of its path. 

Let M be the material point suspend- 
ed from C by the line CM. When re- 
moved from its vertical direction CM 
to the inclined position CM', gravity 
will cause the point to descend and de- 
scribe the arc M'M. Resolving the 
force of gravity g in the directions 
MB and M'T, parallel and perpen- 
dicular to the radius CM', the com- 
ponent g cos. 6, in the direction of CM', 
will be counteracted by the fixed point C. The component g 
sin. 0, in the direction of the tangent M'T, will be the only 
effective force to produce motion. Hence the accelerative 
force is 

f=g sin. 0. 

Let CM=Z, the length of the pendulum, and the arc M'M=s, 
the semi-arc of the vibration; then s=W. 

■ l 
.*• f=g- sin. ft. 




CONSTRAINED MOTION. 183 

Now if x be any arc of a circle, it is shown in trigonometry 
that 

Hence /^(rii/ + l.U4.5.r)' &C * 

But when the arc s is small compared with I, the cube and 

higher powers of j may be neglected, and 

/=f» (85) 

or the force varies as the distance from the lowest point 
measured on the arc. 

301. Prop. When a body is urged toward a fixed point by a 
force varying directly as its distance from that point, the times 
of descent to that point from all distances will be the same. 

For the space described will obviously depend on the in- 
tensity of the force and the time of its action, or will be meas- 
ured by the product of the force by some function of the time. 

Hence s=f<f)(t). 

Since s varies as/, let s=nf n being the constant ratio of the 
space to the force ; then, by substitution, 

n=cp(t) ; 
or the function of the time is the same whatever be the dis- 
tance, and, in the case of the pendulum, n=~. (85). 

302. Schol. It will be shown hereafter that the time of de- 
scent to the center of force, or the point toward which the 



force is directed, is always equal to ~\/— » when \i is the 



ac- 



cr 

celerating force. In the case of the pendulum, i«=y, and we 

shall have, for the time of descent to the lowest point, or the 
time of a semi-vibration, 



184 ' DYNAMICS. 



2V n 



2V'g 

When the material point has reached its lowest position, its 
momentum will cause it to rise on the other side of the vertical 
line ; and, since its velocity at the lowest point will be that due 
to the vertical height fallen through, and the force will require 
the same time to destroy the velocity that was required to 
generate it, the whole time of a vibration will be 



wi 



(86) 
g 

As the circumstances of the material point at the end of the 
time T are the same as at the commencement, it will then per- 
form another vibration in the same time, and so continuing, all 
its vibrations will be isochronal. 

It should be recollected that these results are obtained on 
the supposition that the arcs of vibration are small. 

303. Prop. To discuss the circumstances of the vibration of 
different pendulums. 

1°. Since in (86) rr is constant, 

vV 
or, the time of vibration of~ a pendulum varies directly as the 
square root of the length, and inversely as the square root of the 
accelerating force. 

2°. If g be constant, which is the case in the same latitude, 
and at the same elevation above the surface of the earth, 

T oc VZ; 

or, the time will vary as the square root of the length. 

3°. If / be constant, or the length of the pendulum remain 
the same, then 

Vg 

or, the time of a vibration will vary inversely as the square root 
of the intensity of gravity. 



CONSTRAINED MOTION. 185 

4°. If T be constant, since 

lag; 
or, the lengths of pendulums vibrating in the same time vary as 
the accelerating force. 

304. Cor. Hence the force of gravity in different latitudes, 
and at different elevations above the surface of the earth, will 
vary as the length of the pendulum vibrating seconds. If, 
therefore, the length of the seconds pendulum be ascertained at 
various places on the earth, we shall have the relative intensi- 
ties of gravity at those places. Since these intensities of grav- 
ity at different places are dependent upon the figure of the earth, 
they will serve to determine it. The pendulum, therefore, be- 
comes an instrument for ascertaining the form of the earth. 

305. Prop. The lengths of pendulums at the same place are 
inversely as the square of the number of vibrations performed 
by each in the same time. 

Let N and N' be the number of vibrations performed by 
each respectively in the time H. Then the duration of a vibra- 
tion by each will be 



T:T'=VI:y?=g:| 

Hence 

Also, l=~-j^. (87) 

306. Schol. To determine experimentally the length of the 
seconds pendulum, or a pendulum which will oscillate 86400 
times in a mean solar day, let a clock, the length I of whose 
pendulum is required, be regulated to vibrate seconds. Sus- 
pend in front of the clock a pendulum of known length /'(/'</), 
and observe the exact second when the two simultaneously 
commence a vibration. As the vibration of V will be more 
rapid than that of 7, count the number of returns to coincidence 
of vibration in a period of five or six hours ; and, finally, note 



T= 


=gandT': 


H 


: T'- 


= Vl : 


VI'-- 


H 


1:1'-- 
b 


=N' 3 : 

_/'N' 2 


N 2 . 





186 DYNAMICS. 

the exact second of the simultaneous termination of a vibra- 
tion. If now to the whole number of seconds N shown by the 
clock, there be added twice the number of coincidences of 
vibration, we shall have the number N' of vibrations of /'. 
These values, substituted in (87), will give the length I of the 
seconds pendulum. 

The length of the seconds pendulum in the latitude of New 
York has been determined to be equal to , 

39 in .10168=3 ft .25847. 

307. Prop. To find the value of g, the measure of the intensity 
of gravity. 

From (86), we deduce 

§ rp 2 * 

Making T=l, and using the length of the seconds pendulum 

above, we find, for the value of g in the latitude of New York, 

g-=385 in -.9183=:32 ft .1598. 

308. Prop. To find the correction in the length of a pendulum 
which gains or loses a known number of seconds in a day. 

Let I be the length of the seconds pendulum, n— 86400, 
y= the number of seconds gained or lost in a day, and x= the 
corresponding correction in the length. 

By Art. 305, and recollecting that a diminution of length 
corresponds to an increase in the number of vibrations, 

n 2 : (n+yY=l—x : /, 
from which we deduce 

_ 2ny+y 2 
X ~n 2 +2ny+y 2 ; 

or, rejecting y 2 , as small in comparison with 2ny, 

2y+n ■ 
-~- (88) 

Hence, divide the length of the seconds -pendulum by one plus 



CONSTRAINED MOTION. 187 

the ratio of the number of seconds in a day to twice the gain or 
loss, and the quotient will be the correction in the length. 

x will have the same sign as y. 

309. Prop. To determine the rate of clock when carried to a 
given height above the surface of the earth. 

Let N=86400, N'= the number of vibrations in a day when 
the clock is carried to the height h above the surface of the 
earth, and r= the radius of the earth. The length of the pen- 
dulum remaining the same, 

™ ™ 1 111 
V£ ' V^ N • N'' 

But g : g'=-z : , , 7X2 . 

.-. N : N'=r+h : r, 
and N 



N-N'= 



r+h 

hN 



r+h 
N 
=— r - (89) 

'+! 

N— N', the loss in a day, is the rate of the clock. Hence, 
divide the number of seconds in a day by one plus the ratio of 
the radius of the earth to the height, and, the quotient will be the 
rate. The quantities r and h must be in the same denomina- 
tion. 

310. Cor. If the loss in a day by a seconds pendulum be 
determined by observation, we may find the height to which 
it is carried, for (89) gives 

, r(N-N') , x 

h = W • ( 9 °) 

311. Def. A body suspended by a cord and performing 
revolutions in a horizontal circle is called a conical pendulum. 

312. Prop. To determine the motion of a conical pendulum. 
Let 1= the length of the cord AP, fixed at A and attached 

to the body at P, PC=r the radius of the circle which the body 



188 



DYNAMICS. 



\ /e 




/"" < * 


*».„ 


^ < c 


v 9 





describes with the uniform veloc- 
ity v, m= the mass of the body, 
and the angle PAC=0. Since 
the body is in the same circum- 
stances at each point of its path, 
the forces acting upon it must be in 
equilibrium. These are the ten- 
sion t of the cord in the direction 



mv 



PA, the centrifugal force f=- 
in the direction CP, and the 



weight of the body mg downward. 

Resolving horizontally and vertically, we have 

tsm. 6 =0. 

r 

t cos. 6—mg =0. 
From (6), we have the tension in the cord, 

cos. 6' 
Eliminating t from (a) and (b), 



(a) 
(ft) 



gr 



sin. 



gl 



cos. 
sin. 2 



cos. 
The time t of performing one revolution is 



J== 



2nr 



=^\/- 
=^\/~ 



lcos.6 



g 

AC 

g ' 



313. Dr. Bowditch, in the second volume of his translation 
of the Mecanique Celeste, gives the following formula for com- 
puting the length I of the seconds pendulum in any latitude </>, 

Z=A+a)sin. 2 0, 
in which ^=39.01307 in. and g>=0.20644 in. 



CONSTRAINED MOTION. 189 

314. EXAMPLES. 

Ex. 1. The length of an inclined plane is 400 feet, its height 
250 feet : a body falls from rest from the top of the plane. 
What space will it fall through in 3j seconds? in what time will 
it fall through 300 feet? and what velocity will it have when it 
arrives within 50 feet of the bottom of the plane ? 

Ex. 2. The angle of elevation of a plane is 25°. 30'. A body, 
in falling from the top to the bottom, acquires a velocity of 450 
feet. What is the length of the plane ? 

Ex. 3. The length of a plane is 240 feet, and its elevation is 
36°. Determine that portion of it, equal to its height, which a 
body, in falling down the plane, describes in the same time it 
would fall freely through the height. 

Ex. 4. A body descending vertically draws an equal body, 
25 feet in 2\ seconds, up a plane inclined 30° to the horizon 
by means of a string passing over a pulley at the top of the 
plane. Determine the force of gravity. 

Ex. 5. The time of descent of a body down an inclined 
plane is thrice that down its vertical height. What is the in- 
clination of the plane to the horizon? 

Ex. 6. If a body be projected down a plane inclined at 30° 
to the horizon with a velocity equal to fths of that due to the 
vertical height of the plane, compare the time of descent 
down the plane with that of falling through its height. 

Ex. 7. A given weight P draws another given weight W 
up an inclined plane of known height and length, by means of 
a string parallel to the plane. When and where must P cease 
to act that W may just reach the top of the plane ? 

Ex. 8. Divide a given inclined plane into three parts such 
that the times of descent down them successively may be 
equal. 

Ex. 9. At the instant that a body begins to descend down a 
given inclined plane from the top, another body is projected 
upward from the bottom of the plane with a velocity equal to 



190 



DYNAMICS. 



that acquired in falling down a similar inclined plane n times 
its length. Where will they meet ? 

Ex. 10. A ball having descended to the lowest point of a 
circle through an arc whose chord is a, impels an equal ball 
up an arc whose chord is b. Find e the modulus of elasticity 
of the balls. 

Ex. 11. Determine that point in the hypothenuse of a right- 
angled triangle whose base is parallel to the horizon, from 
which the time of a body's descent in a straight line to the 
right angle may be the least possible. 

Ex. 12. Two bodies fall from two given points in the same 
vertical line down two straight lines to any point of a curve in 
the same time, all the lines being in the same vertical plane. 
Find the equation of the curve. 

Ex. 13. 193 oz. is so distributed at the extremities of a cord 
passing over a pulley, that the more loaded end will descend 
through 3 inches in one second. What is the weight at each 
end of the cord ? 

Ex. 14. If an inelastic body be constrained to move on the 
interior of a regular hexagon, describing the first side with a 
uniform velocity in one second, find the time of describing the 
last side. 

Ex. 15. A stone is whirled round horizontally by a string 
2 yards long. What is the time of one revolution, when 
the tension of the string is 4 times the weight of the stone 1 

Ex. 16. What is the length of a pendulum which oscillates 
twice in one second ? 

Ex. 17. A seconds pendulum, carried to the top of a mount- 
ain, lost 48.6 seconds in a day. What was the height of the 
mountain ? 

Ex. 18. The length of a pendulum vibrating sidereal sec- 
onds, being 38.926 inches, what is the length of the sidereal 
day ? How much must it be lengthened that it may measure 
mean solar time ? 



CHAPTER VI. 




ROTATION OF RIGID BODIES. 

315. Prop. When a rigid body, containing a fixed axis, is 
acted upon by a given force in a plane perpendicular to that 
axis, to determine its motion. 

Let the annexed figure be a section of the 
body by a plane perpendicular to the axis at 
C, and the given force F act at B, in the direc- 
tion AB, at a given distance from C. If v be 
the velocity which the force F can impart to a 
mass M when free, 

F=Mv. 

When the force acts on the body, each parti- 
cle is constrained to move in a circle whose center is in the 
axis through C. If o> be the angle made by the body from the 
action of the force in a unit of time, the linear velocity of any 
particle m will be rco, and its momentum /=mrw. The mo- 
ment of this force, in reference to the axis through C, is 
rf=mr 2 o). The moment of any other particle will have the 
same form, and hence 

rf+r'f'+, &c, =mr 2 G)+m'r' 2 (x)+, &c, 
or 2.rf=2.mr 2 o)=G)'2m?^, 

since w is the same for each particle. 

But the moment of F is rF=Mv.CB, and, as the former is a 
measure of the effect of the latter, 

u2.mr 2 =Mv.CB. 
Mu.CB 



.'. CJ = 



Z.mr 



(91) 



Thus, if the body consisted of two particles m and m' at the 
distances r and r' from the axis, and the force F would impart 
to m, when free, the velocity v, then the angular velocity 



192 



DYNAMICS. 



mvr 



mr*+m'r 
and the linear velocity of m will be 



i„i& 



mvr 



TO): 



mr^+m'r' 2 ' 

If the two particles were equal, and at equal distances from 
the axis, then mr 2 —-m'r' 2 t and 

v 



lfm=m', and r'=2r, 



ra>=; 



v 

ro)=-. 

5 



Cor. If the body were in motion before the force acted on 
it, then the change of angular velocity will be given by (91). 

316. Def. The moment of inertia of a rigid body about an 
axis is the sum of the products of the mass of each particle by 
the square of the distance of that particle from the axis. 

Thus, if m be the mass of a particle of a rigid body, r its 
distance from a fixed straight line, 2.mr 2 is the moment of in- 
ertia of the body about that line. The definition given above 
is to be regarded as a verbal enunciation of this analytical ex- 
pression, which has required nomenclature by the frequency 
of its occurrence in dynamical investigations. 

317. Prop. The moment of inertia of a body about any fixed 
axis exceeds its moment of inertia about a parallel axis passing 
the center of gravity, by the product of the mass into the square 
of the distance between the axes. 

Let m be a particle of the body, and let the 
plane mCG, passing through m, cut the two 
axes at C and G. Through the axis at G 
pass a plane perpendicular to CG, intersect- 
ing the plane of the figure in AB. Let 
mC=r, mG=r^, and CG=h. Draw mF per- 
pendicular and mD parallel to AB. Then 
r*= r \ +h*±2h.GD, 
according as the angle mGC shall be obtuse or acute. Hence 




ROTATION OF RIGID BODIES. 193 

mr 2 =mr\+mh 2 ±2mh.GD, 
and S.mr 2 =S.mrf+A 2 2m±2A2.m.GD. 

Now, since the plane through the axis at G passes through 
the center of gravity, by (29), 

2.m.GI)=2.m.Fm=0. 
.*. ^.mr 2 =I l .mr 2 l -\-h 2 lm 

M being the whole mass of the body. 

318. Schol. Assuming k, such that Mk 2 =2.mr 2 , we have 
2.mr 2 =M(k 2 +h 2 ). 

Mk 2 is the moment of inertia of a body about an axis through 
the center of gravity, and M(k 2 -\-h 2 ) is the moment of inertia 
about a parallel axis at a distance h from the former. The 
moment of inertia about any axis will, therefore, be easily de- 
termined when the moment of inertia about an axis through 

Z.mr 2 
the center of gravity is known. Since k 2 = M l , the determ- 
ination of k will, in general, require the aid of the integral cal- 
culus. 



The length Vk 2 +h 2 is called the radius of gyration about 
the axis considered, and, similarly, k is called the radius of 
gyration about an axis through the center of gravity. Since 
k is the least value of Vk 2 +h 2 , it is sometimes called the prin- 
cipal radius of gyration. 

319. Prop. If a body oscillate about a fixed horizontal axis 
not passing through its center of gravity, there is a point in the 
right line, drawn from the center of gravity perpendicular to the 
axis, whose motion is the same as it would be if the whole mass 
were collected at that point and allowed to vibrate as a pendulum 
about the fixed axis. 

Let the horizontal axis be perpendicular to 
the plane of the figure at C, G be the center 
of gravity, CA be horizontal, and GA vertical. 
Then, if M be the mass of the body, and v the 
velocity which gravity can impart to the body 
if free in each instant of time, the moment of 

N 




194 DYNAMICS. 

gravity will be Mu.CA. By Art. 315, Cor., the change of an- 
gular velocity produced in each instant of time is 

_Mv.CA 
"-^T^- 
Produce CG to some point O and draw OB vertically, meeting 
CA in B. Now, if the whole mass of the body were collected 
at O, the moment of inertia of the body would be M.CO 2 , and 
the change of angular velocity in the same instant of time 
would be 

; _ M^CB 

"'""M.ccr 

But the position of the point O being arbitrary, CO, and there- 
fore CB, will vary at pleasure. We may, therefore, put 
Z.mr* : M.C0 3 =CA : CB=CG : CO, (a) 

which gives 

CA CB 



S.rnr 3 M.CO 3 ' 

or (*) = (*)' ; 

that is, the change of angular velocity is the same in the two 
cases, and, therefore, the motion from rest will be the same. 
320. Cor. The point O may be found from («), which gives 
S.mr 2 M(k 2 +h 2 ) 



CO= 



M.CG - M.h 

k*+h 2 



321. Def. The point is called the center of oscillation of 
the body with respect to the axis through C. It is thus de- 
fined : when a rigid body moves about a fixed horizontal axis 
under the action of gravity, in the straight line drawn through 
the center of gravity perpendicular to the axis, a point can be 
found such that, if the mass of the- body were collected there 
and hung by a thread from the axis, the angular motion of the 
point would, under the same initial circumstances, be the same 
with that of the body, and this point is called the center of 
oscillation of the body with respect to the axis. 



ROTATION OF RIGID BODIES. 195 

322. Cor. Hence, when a body makes small oscillations 

about a fixed horizontal axis, it is only necessary to calculate 

k 2 
the value of the expression A+— =/, and the time of a small 

oscillation will be n\/-. (86). 

© 

323. Def. A body of any form suspended from a fixed axis, 
about which it oscillates by the force of gravity, is called a 
compound pendulum. 

324. Prop. The centers of oscillation and suspension are con- 
vertible. 

Let h' be the distance of the center of oscillation from the 
center of gravity, when h is the distance of the axis from the 
center of gravity. Then, by (92), 

CO-h=h'=^, 
h 

k 2 
and A=— ; (93) 

so that if h' be the distance from the center of gravity to the 
axis, h will be the distance from the center of gravity to the 
center of oscillation, and a body will oscillate in the same time 
about an axis, through the center of oscillation, as it oscillates 
about the original axis, the extent oi vibration being the same. 

325. Cor. 1. If in a straight line through the center of gravity, 
perpendicular to the axis of motion, points be taken at distances 
h and h' from the center of gravity, and on opposite sides of it, 
then the length of the equivalent simple pendulum is h+h' ; so 
that the time of vibration about the axes through each of these 
two points is the same, and the length of the equivalent simple 
pendulum is in each case the distance between the two axes. 
Each of these points is the center of oscillation in reference to 
the other as a center of suspension. 

326. Cor. 2. From (93), we have 

hh'=k 2 ; 
or, the principal radius of gyration is a mean proportional be- 



196 DYNAMICS. 

tween the distances of the centers of oscillation and suspension 
from the center of gravity. 



327. Cor. 3. Since h : Vk*+h 2 = Vtf+h* :h+j=CO, we in- 
fer that the distances of the centers of gravity, of gyration, and of 
oscillation from the axis of motion, are continued proportionals. 

328. Schol. The convertibility of the centers of oscillation 
and suspension was employed by Captain Kater in finding the 
length of a simple pendulum vibrating seconds, and consequent- 
ly the force of gravity at the place of observation. 

A bar of brass one inch and a half wide and one eighth of an 
inch thick was pierced by two holes, through which triangular 
wedges of steel, called knife edges, were inserted, so that the 
pendulum could vibrate on the edge of either of these as an axis, 
resting on two fixed horizontal plates of agate, between which 
the bar was suspended. The axes were about 39 inches apart. 
Weights were attached to the bar and rendered capable of 
small motions by screws, by which means the position of the 
center of gravity of the bar could be changed. These weights 
were so adjusted by trial that the time of a small vibration 
through an angle of about 1° was the same when either knife 
edge was the axis, so that each gave the center of oscillation 
belonging to the other. The distance of the knife edges was ob- 
tained by placing the pendulum so that the edges were viewed 
by two fixed microscopes, each furnished with a micrometer of 
ascertained value, and afterward placing a scale of known ac- 
curacy in a similar position under the same microscopes. This 
method, combined with that referred to in Art. 306, served to 
determine the length of the seconds pendulum, and thence the 
force of gravity. 

329. The relation of the simple to the compound pendulum 
will be illustrated by one or two examples in which the prin- 
cipal radius of gyration is supposed to have been previously 
determined. 

Ex. 1. A material straight line vibrates about an axis per- 
pendicular to its length : required the length of the isochronal 
simple pendulum. 



ROTATION OF RIGID BODIES. 197 

Let 2a be the length of the line, and h the distance of the 
point of suspension from its center of gravity, which is its mid- 
dle point. The radius of gyration of a straight line about an 
axis through its center of gravity, perpendicular to its length, 

a 
is ~7r. Since the radius of gyration is a mean proportional 

between, the distances of the centers of oscillation and of sus- 
pension from the center of gravity (93), 

h'=-=— 
11 h 3h' 

and (Art. 325), l=h+h'=h+^. 

1°. Let the point of suspension be at the extremity of the 
line, in which case h=a. 

Then h'=ia, 

and 7=a+ia = f.2tf, 

that is, the center of oscillation is two thirds of its length below 
the axis of motion. 

2° Let h=\a ; then 

and 7=§.2a, 

the same as before. Hence the time of a small vibration is 
the same, whether the line be suspended from one extremity, 
or from a point one third of its length from the extremity. 
This also illustrates the convertibility of the centers of oscilla- 
tion and of suspension. 
3°. IfA=i« s 

and Z=}fa=f.2a+J.2a; 

or, when the center of suspension is three eighths of its length 
from one end, the center of oscillation is one sixth of its length 
below the other end. 

4°. lfh=0, h'=ao, 

and 1=gd ; 

or, when the center of suspension is at the center of gravity, 
the length of the equivalent simple pendulum is infinite, and 



198 DYNAMICS. 

therefore the time of one vibration is infinite. This is obvious, 
also, from the fact that, when the line is suspended from the 
center of gravity, no motion can result from the action of 
gravity. 

Ex. 2. A sphere being made to oscillate about a given axis ; 
required the length of the equivalent simple pendulum. 

Let r be the radius of the sphere, and h the distance of fhe 
axis of motion from its center. The principal radius of gyra- 
tion of a sphere is rVf. 

2r 2 
Hence &'=— - 

5h 

1°. Let h=r, or the spnere vibrate about an axis tangent to 
its surface. Then 

and l= r +2 r; 

or, the center of oscillation is two fifths of the radius distant 
from the center of the sphere. 



2°. If A=10r, 


r 


and 


l=10r+£. 


3°. If h=}r, 


h'=2r, 


and 


l=2r+-; 



or, the center of oscillation is without the sphere and at a dis- 
tance r from the surface. 



MOMENT OF INERTIA. 



330. Since (Art. 316) the expression for the moment of in- 
ertia of a system consisting of a finite number of points is S.mr 2 , 
m being the portion of the mass which is at the distance r from 
the axis of rotation ; when the number of points becomes in- 
definite, the expression will evidently become fr 2 dM, dM be- 
ing an element of the mass at the distance r. 

Hence (Art. 318), Mk*=fr*dM. (a) 



ROTATION OF RIGID BODIES. 199 

We shall now illustrate the method of determining MF in a 
few simple cases. 

Ex. 1. To find the moment of inertia of a straight line re- 
volving about an axis perpendicular to it at any point of it. 

Let the axis be at a distance a from one end and b from the 
other, and let r be any distance from the axis ; since the thick- 
ness and density of the line are supposed to be uniform, each 
may be taken =1. 

.-. M=a+b and dM=dr. 

r 3 
Hence (a) (a+b)k 2 =f?^dr=—+c, 

3 

and the integral being taken between the limits —a and +6, 

3(a+b)' 

If a= b, or the axis be at the center of gravity, we have, for 
the principal radius of gyration, 

a 

and Mk 2 =---=-a 5 , 

whatever be the thickness and density of the line. 

Ex. 2. To find the moment of inertia of the circumference 
of a circle about an axis perpendicular to its plane through its 
center. 

In this case M=2Trr, and since all the points are at the same 
distance from the axis, r is constant. 

.-. Mk 2 =fr*dM=rydM=2nr\ 
and k=r. 

Ex. 3. To find the moment of inertia of a circle about an 
axis through the center perpendicular to its plane. 

Putting a= the radius of the circle, its area will be 7ra 2 , and 
at a distance r from the axis the area will be irr 2 . Hence, in 
this case, dM=2nrdr and Mk*=na' i k i =27Tfr i, dr=%nr t =%T:a i , 
when r=a, and 



200 DYNAMICS. 

Ex. 4. To find the moment of inertia of the circumference 
of a circle about a diameter. 

Taking the proposed diameter for the axis of z, the distance 
of any point (x, y) from the axis of rotation is y. 

From the equation of the circle, y*=a*—x 2 , we find 



Vdx*-\-dy*=dMz=-dz. 

y 

.-. r*dM=y' i dM.=aydz, 
and Mk*=2'rrak' 2 =afydx—aX area of circle =na z . 

a 

■■ *=7? 

Ex. 5. To find the moment of inertia of a sphere about a 
diameter. 

Let the axis of rotation be the axis of x, and conceive the 
sphere to be generated by a circle of variable radius, y, whose 
center moves along the axis. By Ex. 3 the moment of this 
generating circle is %iry*. Hence the whole moment will be 
the sum of the moments of the generating circle in all its posi- 
tions. 

.-. Mk*=\nfy*dx. 

But y'=a?-x\ .-. Mk^^/^-xydx, 

and this integral, taken between the limits +a and — a, is 
Mk 2 = T %rra\ 

ButM=f7r« 3 . .-. k=ayf\. 



CHAPTER VII. 

331. The methods employed in the preceding chapters to 
determine the motion of a body are partial in their application 
and limited to the simplest cases. When the force is variable, 
or the motion of a body is due to the action of several forces, 
varying, in direction as well as in intensity, with the varying 
position of the body, some more general method is required. 
We now proceed to show, to a limited extent, how the circum- 
stances of the motion of a body or point, in such cases, may be 
determined. In order to this the fundamental formulae will 
need some modification and extension. 

332. By Art. 214, variable velocity at any instant is meas- 
ured by the limit of the ratio of the space to the time, that is, 
from the nature of the differential coefficient, it is the differen- 
tial coefficient of the space regarded as a function of the time. 
Hence 

„=§. rvn-T 

In like manner, it appears from Art. 223, that the force, when 
variable, is measured at any instant by the differential coeffi- 
cient of the velocity regarded as a function of the time. Hence 

*-* [YIIL] 

From [VII. ], by differentiation, we obtain 
dv^cTs 
dt~d?' 

••• *ts*- r ix -] 

From [VII.] and [VIII.] we obtain, by eliminating dt, 
vdu=4>ds. [X.] 



202 DYNAMICS. 

§ I. RECTILINEAR MOTION OF A FREE POINT. 

333. If a free point at rest is acted on by a single force, or 
forces whose resultant is equivalent to a single force, the mo- 
tion must be entirely in the direction of that force, and may at 
once be determined from [VII.] and [VIII.]. But in order 
that we may integrate these expressions, the law of variation 
of the force must be given as a function, either of the space, 
time, or velocity ; and as the intensity of a variable force 
would depend on the position of the body, the force is natu- 
rally and generally given as a function of the distance of the 
body from some fixed point. 

In order that we may obtain an exact expression for the 
value of the force at any variable distance, its value at some 
given distance must be known. This given distance is most 
conveniently assumed as the unit of distance, and if \i be the 
intensity of the force at this unit of distance, it is usually called 
the absolute force. 

If, then, the force be given to vary as the n tk power of the 
distance x from the fixed point, we have 

(j>=[ix\ (94) 

We shall first consider the case of a constant force, for the 
purpose of showing with what facility the calculus enables us to 
determine the relations of the time, space, and velocity, although 
these relations have already been deduced in chapter III. 

334. Prop. To determine the space described by a point acted 
on by a constant force in terms of the time, the velocity of the 
point in the direction of the force at the commencement of the 
time being given. 

The velocity due to the action of the force in the time t is, 
by [III.], 

v'=(f>t f 
where represents the velocity generated by the force in the 
unit of time. 

If Uj be the given velocity, or the velocity when the time 
commences, we have, for the whole velocity, as in (59), 

v=v 1 ±cf>t. 



RECTILINEAR MOTION OF A FREE POINT. 203 

Substituting this value of v in [VII.], we have 

ds=v ^dt^fytdt, 
and, integrating, 

s=vj±\<t>?+c; 

in which c is an arbitrary constant depending on the position 
of the point when 2=0. 

If the space be reckoned from the position of the point when 
t=0, then c=0, and 

s=v 1 t±l(j)t 2 . (60). 
If the point move from rest, v x =0, and 

•=#f. (54). 
All the other relations are readily deduced from these. 

335. Prop. To determine generally the velocity of a 'particle 
moving in a right line to or from a fixed point by the action of 
a variable force. 

By [X.], vdv=(pds ; 

or, measuring the line on the axis of x, 

vdv=<f)dx. 
Hence, by integration, 

v*=2fpdz+c; (95) 

and if the force varies as some function of the distance x, the 
integration may in general be effected. 

If the motion is toward the fixed point, dx will be negative ; 
if from it, positive. 

To determine the constant, we must know the velocity at 
some given point. 

336. Prop. To determine generally the time of the motion. 
From [VII.] we have at once 

dx 
t=f-+c, (96) 

and since v is determined by (95) in terms of x, the integra- 
tion may in general be effected. 

To determine c, we must know the position of the particle 
at some given time. 



204 DYNAMICS. 

337. Prop. To determine the velocity of a particle attracted 
to a fixed point by a force varying directly as the distance of the 
particle from that point. 

Assuming the axis of x in the direction of the motion, and 
the origin at the center of force, by (94) we have 

the sign being negative, because the force which is directed to 
the origin tends to diminish the distance x. 

Substituting this value of </> in (95), and integrating, we get 

v 2 =c— fix*. 

If a be the distance of the particle from the origin when the 
motion commences, when x=a, v=0. Hence 

c=iia 2 , 
and v 2 =[i(a 2 -x 2 ), 97) 

from which it appears that x can not pass the limits ±<z. 

338. Cor. Since the force by which a material point, at 
liberty to move along a perforation from the surface to the 
center of the earth, varies directly as the distance from the 
center, (97) will serve to determine the velocity of the particle 
at any distance x from the center. 

In this case, if a=r, the radius of the earth, and g the force 
of gravity at the surface, ^ will be found from the proportion 
r : ltttgt-p. 

Hence v 2 =-(r 2 -x 2 ). 

As this velocity must be spent before the particle stops, if we 
make v=0, we get x=r, or the particle will go to the opposite 
point of the earth's surface. 

339. Prop. To find the time when the force varies directly as 
the distance. 

dx 
From [VII.], we have dt= — 

1 dx 



vV Va 2 —x 2 ' 
To integrate this, let x—a sin. 0, whence a 2 — x 2 =a 2 cos. 2 6. 



RECTILINEAR MOTION OF A FREE POINT. 205 

1 a.d sin. 6 







y/p a cos. 






= ~.dd, 


and 




t=—=.6+c. 


Hence 




0= Vp(t—c), 


and 




x=a.sin.VfJ<(t—c). 


Suppose 


the time to commence when the particle is at the 


origin, or x 


=0; 


then 

0= — a.sin. V//.C, 


or 




c=0. 
.-. t=-^.d. (a) 



If x=a, sin. 0=1, and d=±n or §7r, &c, 

and t=±n— - or frr— - t &c, 

from the commencement of the motion. 

If x= — a, sin. 6= — 1, and 0=§7r or Jtt, &c, 

1 1 c 

and ^=f7r— = or |ti—=, &c. 

Hence the particle moves from +a to —a, while t changes 
from ^tt^= to !Tr-p=, or the time of one vibration is 

T=Tr-^, (98) 

and this is true whether a be large or small. Compare Art. 
302. 

Cor. 1. When a particle is attracted to a fixed point by a 
force varying directly as the distance, the time of descent to 
that point will be the same for all distances. 

Cor. 2. The time of descent to the center of the earth, in 

which case w=-, is 



206 DYNAMICS. 




=2i , .5 ,/ .a 

Cor. 3. Let S be the center of force, AS=a 
the distance of the particle at the commence- 
ment of motion, P the position of the particle at 
any instant, and t the time of describing PS=#. 
On AS describe a quadrant, and draw PQ, per- 



pendicular to AS. Then 

x=a.s'm. QSB. 
.-. QSB=0andQB=a.0. 



By (97), v= V(*.Va*-x 2 = v>.PQ<xPQ, 

and by («), t =±.8=±M=-r^ * QB. 
J v " Vfi y/\i a vel. atS ^ 

When the particle is at A, arc BQ, becomes BA=|7ra, 
and t=T'=-^=. 

340. Prop. To determine the velocity when a particle is at- 
tracted by a force varying inversely as the square of the distance. 

In this case (94), <f>= fix~ 2 , 

and [X], vdv=—[ix~*dx. 

Integrating, v*= 2fJ,x~ 1 +c. 

If v=0 when x—a, c=—2^a~ l . 

.'. v 2 =2fi(x~ 1 —a~ 1 )=2fi . (99) 



341. Cor. Since the intensity of gravity above the earth's 
surface varies directly as the mass of the body, and inverse- 
ly as the square of its distance from the center, (99) will 
give the velocity of a body falling from any height to the 
surface. When the body arrives at the surface x=r, and 

since \i : g=j^ : —^ fJ>=gr\ Hence 

a— r 

v 2 =2gr. . 

° a 

If a= qo , v= V~2gr= 6.9428 miles, 



RECTILINEAR MOTION OF A FREE POINT. 207 

from which it appears that the velocity can never equal seven 
miles, and if a body be projected upward with the above ve- 
locity, supposing no resistance from the air, it would never re- 
turn. 

If we make x=0 in (99), 

v=ao ; 
or the velocity at the center would be infinite if the same law 
of force continued. 

342. Prop. To find the time when the force varies inversely 
as the square of the distance. 

We have, from [VII.], 

— dx —dx 



dt= =- 



, — fa—x 



ax 
—xdx 
2/x, 



■■ t= (£)Y- 



By adding \adx to the numerator, and then subtracting the 
same quantity from it, we have 

— ( a \ "/*$ \adx—xdx \adx ) 
~\2Ji)J I Vax-x 2 ~ Vax-x 2 > 

= \2ji) K^-^-gversm.- 1 — +c}. 

2x 
When t=0, x—a, and versin. -1 — =ir. 

a 

.'. C = TT, 

/ a \ % x a Qx \ ) 

and t= {-) {(ax-xr+^n-versmr 1 -) J . (100) 

Cor. 1. If x=0, we have 

or the square of the time of falling to the center varies as the 
cube of the distance. 

Cor. 2. On AS=a describe a semicircle, and draw PQ per- 
pendicular to AS. Then PQ= Vax—x\ arc AQS=^tt«, arc 

SQ= SC. L SCQ=^ versin - 1 — . 



208 DYNAMICS. 

Hence the time through AP is, by (100), 
t= (£-\ 2 (PQ+ arc AQS- arc QS) 

= (J)W«cAQ). 
If SQ, be produced to meet in R a tangent at A, 




AR= 



SA.PQ, aVax-x 2 



\ ax J 



PS x 

.'. by (99), v= f^P) 2 . AR x AR. 

When P coincides with S, AR is infinite. 

343. Prop. To find the velocity and time when the force varies 
inversely as the cube of the distance. 

In this case, 0= t*x~ 3 , 

and [X], vdv=—fj,x~ 3 dx. 

Integrating, v 2 = ixx~ 2 +c. 

But when x=a, v=0; .*. c=—fia~ 2 , 
and v 2 = ii(x~ 2 —a~ 2 ), 

_ vV— # 2 
or v=VfJ,. . (101) 

— dr a —xdx 
By [VII.], dt=—=-=. 



v yf\h Va 2 —x 2 ' 



Integrating, 
When x=a, t=0 ; 


f=—=.y/a 2 -x 2 - 

c=0, 
a 


and t=—=.Va 2 —x 2 . 

a 2 
lfz=0, t=~r, 

the time to the center of force. 



(102) 



« ^ , A „ AS.PQ aVa 2 -x 2 
Cor. (See Fig., Art. 339) AR=-p^= . 

.-. v—— i~. AR ac AR, 
a 

and t=— =.PQxPQ. 



CURVILINEAR MOTION OF A FREE POINT. 209 



§ II. CURVILINEAR MOTION OF A FREE POINT. 

344. If several forces in different directions in the same 
plane act continually on a material point, it will have a result- 
ing motion at any instant, whose direction and rate must be de- 
termined by the relations between the directions and intensi- 
ties of the forces which act upon it. The motion may or may 
not be rectilinear, and, in order to investigate the circumstances 
of the resulting motion generally, we shall employ the method 
of resolution to two rectangular axes, as in Statics, using X 
and Y, instead of 2.X and 2.Y, to denote the sums of the re- 
solved forces. 

345. Prop. To determine, generally, the motion of a point act- 
ed upon by any number of forces in the same plane. 

Resolve the forces in the direction of two rectangular axes, 
and let X and Y represent the sums of the resolved forces in 
each axis, x and y being the co-ordinates of the position of the 
point. The point may be regarded as acted upon by the two 
forces X and Y, independently of each other. Hence, by [IX.], 

• d*x w d*y ■ s 

If ds represent any small element of the path of the point, by 

ds 
[VII.], -Ti will represent the velocity of the point in its path. 

Hence, if a and j3 represent the angles which ds makes with 
the axes, the velocities in the direction of the axes (39) are 



ds dx ds - dy 

—.cos. a=— , -£. cos. 0=-tt. (104) 



dt' ' dt 1 dt' dt' 

Such are the general equations by which the motion of a 
point in a plane may be determined. 

346. If the velocity is given, we must, by resolution, obtain 
the velocity in each axis, and thence, by differentiation, the 
force in each axis, and, by composition, the resultant force 
may be determined. 

If the forces are given, we may, by integration, find the ve- 

O 



210 DYNAMICS. 

locity in each axis, and thence, by composition, the velocity of 
the point in its path. 

If the path is required, each of equations (103) must under- 
go two integrations, at each of which operations a constant 
must be introduced. The constants introduced at the first in- 
tegration will depend on the velocity of the point when the 
time commences, or at some given time ; those introduced at 
the second integration will depend on the position of the point. 
We thus have two equations involving the co-ordinates x and 
y and the time t, and, by elimination of t, an equation between 
the co-ordinates will be obtained, which is the equation of the 
path of the point. 

347. Prop. To determine the velocity of a point in its path. 
The velocity may be found as above indicated, but the fol- 
lowing is the preferable method. 

Multiply the first of equations (103) by 2dx, the second by 
2dy, and, adding the results, we have 

2dxd 2 x+2dyd*y ■' ^ r 7 v 
df y * =2(Xdx+Ydy). 

The first member of this equation is the differential of 
dx 2 +dy* _ds*_ 2 
~~df ~df~ V ' 
Therefore, integrating, we find 

v 2 =2f(Xdx+Ydy)+c. (105) 

348. Cor. If this expression Xdx+Ydy is integrable, the ve- 
locity may be found, provided we can correct the integral, or 
know the velocity at some given point. 

Thus, if the expression is a differential of the co-ordinates 
of the position of the point, so that its integral is a function of 
these co-ordinates, or 

v*=2f(x,y)+c, 
and if tj j is the velocity at the point whose co-ordinates are a 
and &, so that 

«;=2/(a, b)+c; 
then tf-v\=2f{x, y)-2f{a, b). 

Hence it appears that the velocity acquired by the particle 



CURVILINEAR MOTION OF A FREE POINT. 211 

in passing from one point (a, b) to any other (x, y) is the 
same, whatever be the curve described between these points, 
since the change of velocity is independent of the co-ordinates 
of any intermediate point. 

349. Prop. The expression Xdx-fYdy is always integrable 
whenever the forces are directed to fixed centers, and their intens- 
ity is a function of the distance from those centers. 

Let F, F l . . . be two forces directed to fixed centers, 
" a,b 9 a l ,b l be the co-ordinates of the centers, 
" x, y .... be the co-ordinates of any position of the 

particle, 
" r,r^.. be the distances of the particle from the cen- 
ters, 
" a, 0, a lf (3 X be the angles which r and r, make with 
the axes. 
Resolving F parallel to each axis, we have for the com- 
ponents 

F.cos. a, F cos. (3. 

But cos. a— and cos. B=- — . 

r r 

Hence the components are 

r ' r 
The other forces may be resolved similarly. 
Therefore the expression 

Xdx+Ydy=F ( X ^dx+^) 






But r » = ( a; _ a )» + ( y _6)» > 

x~~~ a v , ~ — b 

and, by differentiation, dr= dx+- dy. 

r r 

x~~ a i/~~~ b 

Similarly, dr x = l dx+~ — -dy. 

.\ Xdx+Ydy=Fdr+F l dr 1 . ' 
And since, by hypothesis, each force F is a function of the dis- 



212 



DYNAMICS. 



tance r, each of the terms of the second member of the equa- 
tion is an exact differential, and therefore the first is also. 

In the same manner, the reasoning may be extended to any 
number of centers of force. 

350. Prop. To determine the motion of a point acted upon by 
a force directed to a fixed center. 

Assuming the fixed point as the origin of co-ordinates, let F 
be the force, whether a single one or the resultant of several, 
and r the distance of any position of the particle from the cen- 
ter, called the radius vector. 

Multiply the first of equations (103) by y, and the second 
by x, and subtract the former from the latter. This gives 

^9^=,Y-,X. (!06) 

Resolving F in the direction of the axis, we 

have X=F-, Y=F^. 

r r 

Multiplying the first by y, and the second 

by x, and taking the difference, we get 

xY-yX=0. 

Therefore the first member of (106) is zero, and since it is the 

xdv — xi dx 
differential of — , by integration we obtain 




/ 



xd 2 y —yd*x_ xdy — ydx _ 

~~ar ~ It ~ c ' 



or xdy—ydx=cdt. (107) 

Let o) be the angle made by r with the axis of x ; then 

x=r cos. W, (a) 

y=r sin. co ; (b) 

also, dx— cos. a) dr—r. sin. o). do>, (c) 

dy= sin. odr+r cos. a>.dw. (d) 

Multiplying (c) by (b), and (d) by («), and subtracting, we 
get 

xdy — ydx = r 2 do) = cdt. 
r 2 d(*) 



CURVILINEAR MOTION OF A FREE POINT. 213 

But |r 2 do) is the differential of the area of a plane curve re- 
ferred to polar co-ordinates. Hence the ratio of the element 
of the area to the element of the time is constant, and the area 
described by the radius vector in any time is proportional to the 
time. 

By integrating \r*d<>)=\cdt, we have for the area described 
in any time 

k=\ct, (109) 

and if t=\, we have c=2A, 

or the constant c is twice the area described in the unit of 
time. 

351. Prop. Conversely, if a material point describe by its 
radius vector around a fixed center areas proportional to the 
times, the force is directed to that point. 

When the radius vector describes areas proportional to the 
times, we have (107) 

xdy—ydx 
Jt =C ' 
Differentiating, we get 

xd*y— yd 2 x 

df =0 " 

.-. (106) xY-yX=0, 

and X : Y=x : y. 

Hence the forces X and Y, which are parallel respectively 
to the co-ordinates x and y of the point P, are proportional to 
these co-ordinates, and the resultant PR of X and Y must take 
the direction of PS, which is the hypothenuse of the right-an- 
gled triangle PMS, in which PM=y and SM =a;. 

As an application of the general equations of motion of a free 
point (103), we may take the following cases. 

352. Prop. To determine the motion of a point moving from 
the action of an impulsive force. 

Since, after the action of the impulse, the point is abandoned 
to itself, there are no accelerating forces acting on the point, 
and we have (103) 

d^x i_ d 2 y , r 




214 DYNAMICS. 

t dx dy 

Integrating, ST^' fa" 2 ; (a) 

v, and v 2 being constants, added to complete the integral. 

Compounding these velocities (Art. 236), we have for the 
resultant velocity 



v=y/- 



dx 2 +dy* ds 



=^-V¥f+^. 



df dt 

The velocity is therefore uniform. 
Integrating equations (a), 

x=v y t+s x , y=v 2 t+s 2 , (b) 

the constants s x and s 2 being the co-ordinates of the point when 
the time commences. 

Eliminating t from equations (b), we have 

y=-JL x+V2 (s 2 - Si ), 

the equation of a straight line. 

Hence the path of the point is a straight line and the motion 
uniform, which accords with the first law of motion. 

353. Prop. To determine the motion of a projectile acted upon 
by gravity regained as a constant force. 

In this case, taking the axis of x horizontal and that of y ver- 
tical, (103) give 

d*x ^ r d 2 y ^ T 

Multiplying by dt, and integrating, 
dx dy 

Multiplying by dt, and integrating again, we have 

x=v l t+s 1 , y=v 2 t-±g?+s 2 . (a) 

But if the point of projection be at the origin of co-ordinates, 
and t be reckoned from the commencement of motion, s l and s z 
will each be zero. Putting v 2 =av. lt s x =0, s 2 =0, and elimi- 
nating t from the equations (a), we get 

y=ax—x i —- 2 -. 

Compare this with (63). 



CONSTRAINED MOTION OF A POINT. 



215 



III. 



CONSTRAINED MOTION OF A POINT. 



354. Prop. To determine the velocity of a point moving on a 
given curve. 

If a point is constrained to move on a curve, the reaction of 
the curve will be a new force normal to the curve, and may- 
be resolved in the direction of the axes and combined with the 
other components, as in Statics, Art. 77. 

Let N be this normal force, a and (3 the angles which the 
normal makes with the axes, and X and Y the sums of the 
components of all the other forces in the direction of the axes. 

Resolving N in the same directions, we shall have from (103) 
aVx 
~d? 



9 -=X+N cos. a 



S^Y+Nco.,.,3 



But cos. a= sin. Pti 




H M 



(a) 

(P) 



dy 

*"ds' su PP osm & 

the motion in the direction iP, and cos. 

dx 
0=sin. a=cos. J ) tm= — ~ r . 
as 

dt ds 

dt ds' 

Multiplying (a) by 2dx, and (b) by 2dy, and adding 

2dxd*x+2dyd 2 y rt ■ 7 .■__,,. 
df y J =2(Xdx+Ydy). 

The first member is the differential of 
dx*+dy 2 _ds 2 _ a 
~~d^'~df~ V ' 
Whence, integrating, 

v*=2f(Xdx+Ydy)+c. (110) 

Cor. 1. Since this result is exactly the same as that obtained 
for unrestrained motion (105), we may conclude, 



216 DYNAMICS. 

1°. That, if no accelerating force act on the point, or X=0 
and Y=0, its velocity will remain constant, and be not at all 
retarded by the action of the curve. See Art. 284. 

2°. If any accelerating forces do act on the point, the veloc- 
ity is independent of the curve on which it is constrained to 
move. See Art. 286. 

Cor. 2. If the accelerating forces are all parallel, they may 
be assumed parallel to the axis of x, and in this case (110) be- 
comes 

v*=2fXdx+c. (Ill) 

Cor. 3. If in this last case the force, or resultant of the forces, 
is constant, and equal to f, we have 

v 2 =±2fx+c, 
where the upper or lower sign is to be used according as the 
force tends to increase or diminish x. 

If the distance of the point from the origin, x=a, when u=0, 
c=2fa, and 

v*=2f(a-x); (112) 

and since only the ordinates on the axis of x are involved, the 
velocity of the point on the curve depends not on the curve de- 
scribed, but on the difference of the ordinates on the axis of a;. 

355. Prop. To find the time of motion of a point on a given 

curve. 

ds 
In all cases [VII. "I, dt=—. 

Hence, when the nature of the curve and the velocity at any 
point of it is known, the value of v being found from (110) and 
substituted in that of dt, the time may be found by integration. 
Cor. If the forces act in parallel lines, and their resultant is 
constant, by (112) 

dt= dS . (113) 

V2f(a-x) 

356. Prop. To find the reaction of the curve. 

dy dx 

Multiplying (a), Art. 354, by — , and (b) by — , and subtract- 
ing the latter from the former, we get 



CONSTRAINED MOTION OF A POINT. 217 

N=Y dx x chj i d y d * x ~ dxd *y 

ds ds dsdf 

Eliminating dt by the equation v=-j-, 

ds ds ds 

But if p be the radius of the osculating circle, 

ds 3 
dyd 2 x—dxd 2 y 
__ ^ T dx ^ r dy v 2 
ds ds p 

v 2 . 
=Y cos. (3— X cos. oH — 
P 

The first two terms of this value of N are equal and oppo- 
site to the forces X and Y resolved in the direction of the nor- 
mal. They give, therefore, the pressure on the curve due to 

v 2 
the action of these forces, while the other term — is the reac- 

9 
tion of the curve or pressure due to the motion. 

Otherwise. If we suppose no accelerating forces to act on 
the point, then X and Y are each equal to zero in (a) and (b), 
Art. 354. Hence 

(dx>+dtf\ _K,_ (d'x)'+(d'yy 

* \ ds' J ~ dt 

ds 

or, eliminating dt by the equation dt=— 9 



V(d 2 x) 2 + (d 2 yY 
ds 2 
But the coefficient of v 2 is the reciprocal of the radius of 
curvature when s is the independent variable. 

•■■ N=~. (114) 

Compare this with (79). 

As an application of the formulae of constrained motion, we 
may take the following cases : 



218 DYNAMICS. 

357. Prop. To determine the motion of a body descending by 
the force of gravity down the arc of a vertical circle, 

1°. To find the velocity. Formula (112) applies to this case, 
and if A=SA be the height from which the 



a / body at P descends, and x any distance from 
y the lowest point on S A, the axis of x, we have 



s v= V2g(h—x). 

When x=0, we have for the velocity at the lowest point, 

v= V2gh, 

which is the same as that due to the vertical height h. 

2°. To find the time, we have from (113) 

ds 
dt=- 



V2g(h-x) 
The equation of the circle is 

y 2 =2ax— x 2 . 
(a—x) 2 dx 2 



Hence dy 2 



2ax- 



But ds*=dx 2 +dy 2 =dx 2 (l+ j. a ^', ) 

/tax — x / 



a 2 dx' 



ds= 



2ax-x* 
adx 



V2ax—x 2 

This, taken negative, because the arc is a decreasing func- 
tion of the time, and substituted in the above, gives 
a dx 



dt=- 



V2g V(h-x) (2ax-x 2 ) 
a dx 

V~2g' V(hx-x 2 ) (2a-x) 

a (2a—x)~2dx 
V~2g' Vhx—x 2 



(a) 






c-s) ** 



g Vhx-x" 



CONSTRAINED MOTION OF A POINT. 219 

Expanding (1 — — ) by the binomial theorem, we have 
1 Fa dx {i If x\ ISfxy 135fx\\ 

&c.J. 

x n dx 

Thus the terms to be integrated are of the form 



Vhx— x* 

the exponents n being natural numbers beginning with 0. Per- 
forming the operations and taking the integral between the 
limits x—h and x=0, we find 

If h be small, the first term will give an approximate value 
for the time, viz.. 



=*vi- 



Otherwise. If in (a) x be rejected as small in comparison 
with 2a, the equation reduces to 

, /a <& 



\ t=— i\/ — .versin. -r+c. 
g h 



g' Vhx—x 
But t=0 when x=h; 

••• c =^v J 

, /a . . _,2x\ 

and i —i\/~'( n ~~ versin. -jr), 



and when x=0, t=\n\J-. (See (86) ). 

o 

358. Prop. To determine the motion of a body descending by 
gravity down the inverted arc of a cycloid whose base is hori- 
zontal and axis vertical. 

The cycloid being inverted, let the origin be at the lowest 



220 



DYNAMICS. 



point, the axis of x being vertical, and that of y horizontal 
Let the axis AB=2«, AM=a;, PM=y, and AP=s. 

The velocity, as in the preceding propo- 
~J sition, is 




h being the abscissa of the point K from 
which the body begins to descend. 

The differential equation of the curve, referred to the vertex 
as the origin, is 

(2ax— x)dx 



dy— 



V2ax—x' i 



But ds>=df+dx>= (g-jg + i)&>. 



.*. ds=dx\/ — 
V x 



Substituting this value of ds with its proper sign in [VII.], 
we find 



dt\ 



Vi 



I ' a 2x\ 

Integrating, t = \J — . (n 



dx 
•x— 

versin. -1 -^ ] 



g' Vhx—x 



t^V-AV^ 



and therefore for the time of descent to the lowest point, where 
x=0, 

'la 

g 2V 7" 

Cor. The time t being independent of the abscissa h f will be 
the same, from whatever point in the curve the body begins to 
descend ; in this respect differing from the motion in circular 
arcs. On account of this remarkable property the cycloid is 
called tautochronal. 

359. Schol. The property of tautochronism which attaches 
to the cycloid was supposed to give peculiar advantages to a 
pendulum made to oscillate in this curve. The method of ac- 
complishing this is naturally suggested by the property of the 
evolutes ; and as an evolute of a cycloid is another equal cy- 



CENTRAL FORCE. 221 

cloid, we have only to take two equal semicycloids and place 
the extremities of their bases contiguous in 
the same horizontal line as in the figure. 

If we then suspend a body at the point S, b s 

common to the two semicycloids, by a flexi- 
ble string equal in length to either semicy- X^ 
cloid, or, which is the same thing, to twice A 

the axis, and make it oscillate between the two, it will generate 
the involute, or another equal cycloid. 

The constant change, however, in the center of motion 
arising from the contact of the string with the two curves or 
checks, which changes also the relative velocity of the different 
parts of the vibrating body, renders this contrivance, although 
beautiful in theory, yet useless in practice ; independently of 
the difficulty of obtaining a string sufficiently flexible, and of 
ensuring accuracy in the plates. 

Vibrations, therefore, in small circular arcs, which are at the 
same time also most natural, have been adhered to in practice, 
and it has been shown (Art. 302 and 300) that, as long as the 
arc is small, these vibrations have all the advantages of the 
vibrations in cycloidal arcs. 



§ IV. MOTION OF A POINT ACTED UPON BY A CENTRAL FORCE. 

360. The case of a point revolving in an orbit by the action 
of a force tending to a fixed center is of sufficient importance 
to justify a distinct discussion, especially as the formulae are 
susceptible of considerable simplification, when the force is 
given as a function of the distance from the center. The fun- 
damental equations 

are components of the force in two rectangular directions 
But when the force is directed to or from a fixed center, and 
is given in functions of the distance of the particle from that 
center, it is most natural and convenient to introduce polar co- 
ordinates. 




222 DYNAMICS. 

361. Prop. To transform the expressions for the force from 
rectangular to polar co-ordinates. 

Suppose the origin to be at S, the center of 
force, and let r=SP, the radius vector or dis- 
% r ^ < \_ tance of the particle, and w==PSX, the angle 
made by r with axis of x. 

Then X= — F cos. u, Y= — F sin. w, 
the signs being negative because the force is supposed to be 
attractive, or act toward the center in opposition to X and Y. 
Multiplying the first by sin. cj, and the second by cos. w, and 
subtracting, we have 

X sin. 6)— Y cos. a)=0. (a) 

Similarly, multiplying the first by cos. w, and the second by 
sin. w, and adding, we get 

X cos. w+Y sin. *>=— F. (b) 

But x and y being the co-ordinates of the position of the 
particle 

x=r cos. w, y=r sin. o>. 
Differentiating each twice, we obtain 

d 2 x=d 2 r.cos. o)—2dr.do).s'm. o)—rdo) 2 .cos.o)—rd 2 G).sin. w, 
d 2 y=d 2 r. sin. cj-\-2dr.d(o.cos.(*)— rdo) 2 .s'm. o+rd 2 (*).cos.G). 

By dividing these expressions by df, the values of X and Y 
will be obtained (103), and if we substitute these values in (a) 
and (6), we get 

drdo) d 2 (*> ,„, . 

d 2 r d(*) 2 „ 

__ r _ +F=0( (U 7) 

which are two polar equations to determine the motion. 

362. Cor. 1. The radius vector of a point involving about a 
center of force describes equal areas in equal times. 

For, multiplying (116) by rdt, we "obtain 
, do) n d 2 o , r 2 d(o 



CENTRAL FORCES. 223 

Integrating, ~~JT~ Cl 0*8) 

and" fr'd^—ct-\-c x . 

But \fr 1 du is the area described by the radius vector in the 
time t, an d it varies as t. 

363. Cor. 2. The angular velocity of the radius vector varies 
inversely as the square of the distance of the particle. 

du c 
From (118) we obtain ^=-- (U9) 

But -37 is the angular velocity of ?•, and it varies inversely 
as r 2 . 

364. Prop. To determine the velocity of the particle in its 
orbit. 

Multiply (116) by rdu, and (117) by dr, and add the results. 
This gives 

d*r do? d*o) 

, (dr' do) 2 \ __ , 

d \w^w) +2Fdr =°- 

Therefore, integrating, we have 

^ +2fFdr=c>. (120) 

, ds* dr'+r'do) 2 /Ci 

But v df = — If * ^ ' 4 ^ 

... »W-2/*Ftfr. (121) 

Cor. Since the value of v depends only on the distance of 
the particle from the center, the velocity of the particle will be 
the same at any two points equally distant from the center, and 
the velocity acquired in passing from one point to another will 
be the same, whatever be the path or curve between them, if 
the law of force remain the same. 

365. Prop. To find the time of describing any portion of the 
orbit. 



224 DYNAMICS. 

By (ii 9)> §4 

which, substituted in (120), gives 

dr* c 2 

— +?+2 fFdr=c>. (122) 

In this expression we have the relation of r to t, independent 
of the orbit. 



366. Prop. To find the equation of the orbit. 
Substituting 
(118), we get 



Substituting in (122) for df its value — — , obtained from 



(^ + ?) +2 ^ r = c ' ; < 123 > 



or, to give the equation a more convenient form, put r=—_ \ from 

du 

which we find dr— r. These values of r and dr, substituted 

u 2 



in (123), give 



*1&*-#&*. <«*> 



When the law of force is given, we can obtain from this ex- 
pression, by integration, the relation between u and w, or the 
polar equation of the orbit ; or, if the relation between u and w 
is given, the law of force may be determined. 

367. Schol. A familiarity with the geometrical representa- 
tives of the terms and factors in the foregoing formulae will 
conduce to facility in their application. 

In the annexed figure let the 
%^ center of force be at S, the origin 

of co-ordinates, SP=r the radius 
vector of the particle at P, W=ds 
x an element of its path coinciding 
a"s with the tangent PT, w=PST the 

angle made by the radius vector with the axis of x, V ; SV=d(o 
the angle described by the radius vector in the indefinitely 




CENTRAL FORCES. 225 

small time dt, and mY'=dr the increment or decrement of the 
radius vector in the same time. Let the arc Ym be described 
with S as a center, and radius SP, and the arc nn' with the 
radius Sw=l. Then, 

1°. Since Sn : &Y=nn' : Ym or 1 : r=du : Ym, 

Ym=rd(o. 
2°. The area of the sector SPm is 

A = — - — == lr 2 d(o. 
2 2 

3°. If the point P' be indefinitely near to P, PP' and Ym may- 
be considered straight lines, and the triangle YmY' a plane 
rectilinear triangle, right-angled at m. Hence 
PP=P™ 2 +Pm 2 , 
ds 2 =rdrf+dr\ 
4°. If Sp^p be a perpendicular on the tangent, when the 
triangle YmY' is in its nascent or evanescent state, the angle 
YY'm=pYS and the triangle PP'??i is similar to pYS. Hence 

PP : Ym=YS : Sp, or ds : rdu=r : p. .-. p=^, and [VII.], 

r 2 d(o r 2 d(o , _ „ _ 

p= — j-. .'. vp=—j—= 9 by (118), twice the area described 

in the unit of time. 



AIs0 ' ^-"^-^+rW 

5°. Since Sp : pY : YS=Ym : mP : Y'Y=rda : dr : ds 9 
Sp Ym rdu . 

■'■ ^ = PP = ~5" =sm " F ' 

Yp _Y'm_dr _ 

ps~pp~ ds ~" cos • * 

»S mY rdo) _ 

and ^_:=— ;=-— =tan. P. 

jor mr' dr 

6°. If the force F be resolved in the direction of the curve, 

we shall have 

d 2 s „ „ ^dr 
_=-Fco,P=-F^. 



226 DYNAMICS. 

Multiplying both sides by 2ds, and integrating, we have 

ds 2 

—= v *=c'-2fFdr, as before (121). 

By reference to (117), we see that the first term has the 

form of the differential expression [IX.] of a force in the direc- 

r .i i rdo)2 r'dG? _ rdu . . . 

tion oi r ; the second term —r^-= — rr. But —7- is the arc of 

df rdf dt 

a circle of radius r, divided by the time. It is, therefore, the 

i- i • r i , ■ • i tt r 9 d« a v 2 . 

linear velocity v 1 of a body in a circle. Hence , a =—=f, 

d 2 r 

the centrifugal force in a circle. .*. (117) becomes -r T =f—¥, 

F being the force by which the point describes any orbit, and 

/ the centrifugal force in a circle whose radius is r, the dis- 

d 2 r 
tance of the point at the instant, -77, the difference of these 

forces at any time, is that by which the radius vector is in- 
creased or diminished, and is called the paracentric force. In 

d 2 r dr 
like manner, the integral of -77 or -77 is the velocity of ap- 
proach or recession from the center, and is called the para- 
centric velocity. The paracentric force is the difference be- 
tween the centrifugal and centripetal forces. 

_ _,. rdu 2 1 r 4 dw 2 c 2 

Co*. I. Since —=-.—=-. 

■■•/=$■■ (125) 

Cor. 2. The quantity by which c 2 is multiplied in (123) is, 
by (4°), equal to — . Hence 

j 2 +2f¥dr=c>. 

Differentiating, p being variable, 

2c 2 dp _ , 
+ +2Fdr=0. 



CENTRAL FORCES. 227 

Hence F=|g, (126) 

an expression most useful in finding the law of force by which 
a given orbit may be described. 

ds 2 dr*+r*d(S r'dco 2 c 2 



Cor. 3. Since v 2 = 



dt df fdt f 



pdr 
But —j-=\ chord of curvature through the pole, and (if F 

be constant) is (55) the space through which a body must fall 
to acquire the velocity v. Hence the velocity of a body at any 
point of its orbit is equal to that acquired by falling through 
one fourth the chord of curvature through the pole. 

Application of the preceding formulas to the motions of the 
planets. 

368. Observation has established three facts respecting the 
motions of the planets, which, from their discoverer, are called 
Kepler's Laws. 

1°. The areas described by the radius vector of a planet are 
proportional to the times. 

2°. The orbit of a planet is an ellipse of which the center of 
the sun is one of the foci. 

3°. The squares of the times of revolution of the different 
planets are proportional to the cubes of their mean distances 
from the sun, or the semi-major axes of their orbits. 

These laws relate only to the center of inertia of each planet, 
and conclusions from them must be limited to the motion of 
that point ; in other words, to the motion of translation of the 
planets. 

369. Prop. The accelerating force by which a planet describes 
its orbit is directed to the center of the sun. 

For, by the first of Kepler's laws, the areas described by 
the radius vector are proportional to the times, and when this 
is the case, by Art. 351, the force must be directed to the point 
about which this equable description of areas takes place. 



228 DYNAMICS. 

370. Prop. The force by which a planet describes its orbit 
varies inversely as the square of its distance from the center. 

By the second law the orbit is an ellipse. The equation of 
an ellipse referred to the focus as a pole is, 

«(l-e 2 ) 

r= ? -r— - r— 5T, (128) 

1-fecos. (a>— 6) v ' 

in which a is the semi-major axis, called the mean distance, and 
e the eccentricity. The angle 6 is an angle 
^b made by a fixed line, through the focus with 




the major axis, reckoned from the lower 
apsis, or least distance, and called the longi- 
tude of the perihelion, the lower apsis being called the perihel- 
ion, and the higher the aphelion. The angle (o—6, which ex- 
presses the angular distance from the perihelion, is called the 
true anomaly. 

Putting r=- in the above equation, we get 

1+e cos. ((o—d) 

a(\ — e 2 ) v ' 

Differentiating in reference to (o, 

du e.sin. ((o—0) 
d(o~ a{\— e 2 ) 
Differentiating, again, 

d 2 u_ e cos. ((0—6) 
dtf~ a(l-e 2 ) * 
Adding (b) to (a), 



© 



d(o* a(l-e*) 

Differentiating (124), 

d(o^ U ~&u 



u 



Hence F=— - ^=~n iv - * (129) 

a(l— e 2 ) a(l— e 2 ) r a v ' 

The coefficient of — being constant, the force varies inverse- 
ly as the square of the distance. 



CENTRAL FORCES. 229 



Hence the coefficient of -5 is the intensity of the force at the 
unit of distance, or the absolute force. 

371. Prop. The intensity of the force is the same for all the 
planets at the same distance. 

The quantity c (109) is twice the area described in the unit 
of time. If, therefore, T be the time of one revolution of a 
planet, cT will be twice the area of the ellipse described by 
the radius vector. 

But the area of an ellipse is nab=nd t Vl—e i . 

and 



.-. cT=2naWl-e 2 , 
c* 4rrV 



a(l-e*) T * 
Similarly, if a', e 1 , c', T' express the same quantities in any 
other orbit, we shall have 

c' 2 4rrV 3 





a'{\- 


-e") 


FT1/2 • 


Kepler's 


third law, 








rpa . 


T /2 -_ 


a 3 : a'% 




C 


2 


c n 



" a(l-e 2 ) a'(l-€ ,2 Y 

Hence the force does not vary from one planet to another 
except in consequence of change of distance. 

372. Prop. To find the velocity of a planet at any point in its 
orbit. 

By (121), v*=c'-2fFdr. 

Substituting in this the value of F (129), we have 
a _ 2c 2 rdr 

V ~ C a(l-e 2 )'J re- 
integrating, ^^—X 

To determine c', we must know the velocity at some given 



230 



DYNAMICS. 



distance, or at what distance and with what velocity the planet 
was originally projected into space. 

Let v l be this primitive velocity, and r, the corresponding 
distance. 

Then v 2 =c>+-f- —±, 

1 a(l— e 2 ) r a 

. „ 2c 2 1 
or c 



1 a(l-e")V," 

1 «(l-e 2 )' 



-(J-^-). (130) 



i 
Otherwise. From Art. 367, Cor. 3, we have 

5 2 r a 2 (l—e 2 ) 
and (Analytic Geometry) p*=^—^= ^_ ? 

c 2 2a— r 



(131) 



a 2 (l-e 2 )* r * 

Cor. 1. Hence the velocity is greatest when r is least, or at 
perihelion, and least at aphelion, where r is the greatest. 

Cor. 2. If a body move in a circle whose radius is r, v Y be- 
ing its velocity, the central force is 
v 2 c 2 1 

r a(l— e 2 ) r 2 v ' 
if the force is the same as that which retains the planet in its orbit. 

■ ■. » = ° 2 2a ~ r - c * * 
'"" U : Vl « 2 (l-e 2 )* r : tf(l-e 2 )V 

=2a— r : a. 

That is, the square of the velocity in the ellipse is to the square 

of the velocity in the circle as the distance of the planet from 

the unoccupied focus is to the semi-major axis. 

Cor. 3. If r j =perihelion distance, and r 2 =aphelion distance ; 

c 2 2# — ~r c 2 r 

then, if in (131) r=r„ ^,=^^.—-^=-^--^.-1, («) 

— 2 _ & 2a—r 2 _ c 2 r, 
r ~ r 2> v ~u\\-^)' r* - a Xl-e*)'r~ 2 ' ( b ' 

c 2 



r=a > v -^wT v ^-® ; 



CENTRAL FORCES. 231 

or, the velocity at the extremity of the minor axis is a mean 
proportional between the velocities at the perihelion and 
aphelion. 

373. Prop. To determine the orbit which a body will describe 
when the force varies inversely as the square of the distance. 

If \h be the force at the unit of distance, then 

r 

.-. f¥^=f\idu=iLu. 
This, substituted in (124), gives 

c 2 \j~i+u 2 )-2^u=c f J 

which is the differential equation of the orbit 
To integrate this, we find 

_ _ c'du' 2 

dd)-: 



2„,2* 



c'+2im—cu 

\i 1 

Put u=z+— , and b~=— (c'c'+fj,*), and we have 
c c 

dz 
du=- 



Vb 2 -z- 

z 
or, integrating, w=0+cos. l -. 

.'. z=b cos. (o) — d) ; 
or, replacing the value of z, 

u=^+b cos. (to— 0). 



Hence r= t-j-^ , (132) 

H cos. (cj— 6) 

which is the equation of a conic section referred to the focus. 
Hence the orbit which a body will describe when the force 
varies inversely as the square of the distance is a conic section. 
To determine the constants, and therefore the dimensions 
and form of the orbit, some other circumstances must be given. 



232 DYNAMICS. 

Suppose we know v 1 the initial velocity or velocity of pro- 
jection into space, p the distance of the planet from the sun, 
and if) the angle made by the direction of projection with the 
distance. 

By Art. 367, 5°, sin. i>= 9 -^-. 

as 

tt / «v P 2 ^ W ds pdto 

Hence (118), c=— =p.j t -^ =P-v. sm. +. (a) 

By (121), v'=c'-2fFdr=c'+— . 



© 



.'. c'=-v 2 . . 

p 

Comparing (132) with (128), we see that 

a(l-e^ and e J^=-(c'c^^f. 

•• a =-c>=2^P {C) 

c , c * p a v*8mS1>(v>~) 
and e 2 =— +1 = 1 + 5 v — . (d) 

Cor. From (c) it appears that the semi-major axis « depends 
only on the distance p and the velocity of projection v l9 and 
is independent of the angle of projection ip. Hence, in what- 
ever direction the body is projected, the major axis of the orbit 
will be the same. 

The orbit will be an ellipse, hyperbola, or parabola, accord- 
ing as the value of e is less than, greater than, or equal to unity, 
or (d) according as 



2fi 



P 
is negative, positive, or equal to zero. 

Or, since (99), v 2 = 2f *(l--), if a=ao, v'=^. Therefore, 

when a body is attracted from an infinite distance to a center 
of force which varies inversely as the square of the distance, 
the square of the velocity at any distance p is 



CENTRAL FORCES. 233 

v:= — and v: =0. 

P P 

Hence the orbit will be an ellipse, parabola, or hyperbola, ac- 
cording as the velocity of projection is less than, equal to, or 
greater than, that acquired from an infinite distance. 

374. Prop. To find the time of describing any portion of the 
orbit. 

Substituting the value of F=— in (122), we get 

d? + 7~~r~~ C ' 
The values of the constants found in terms of the elements 
of the orbit in the preceding proposition are 

u 

c'= — and c 2 =fia(l— e 2 ). 

r^d 
Substituting these values, and multiplying by — , we obtain 

a r^dr* 

\i df v ' 

/ d\ 2 rdr 

Hence dt= ( - ) . , 

W vVe 2 -(a-r) 2 

In order to integrate this expression, let {a—r)=aez, then 

. a.\ 2 (1— ez)dz 
dt=-a' 



H>S Vl-z* 
_/a 3 \2 / —dz zdz i 

>/*/ ' \ vi-% 2 e vT^)' 

( 3 \ - 
— ) jcos. _1 % — e(l — z^l+k. 

When 25=1, or r—a—ae, and the body is at the nearer apsis, 
then t 1 =k= the time at the perihelion. 

Hence the time through any portion of the arc from the 
perihelion is 

*-*,= (— ) 2 {cos.- 1 z-e(l-z 2 )^. (133) 



234 DYNAMICS. 

If z= — 1, r=a-\-ae, and the body is at the further apsis, in 
which case t—tj is the time of half a revolution, and 

Also, if T be the periodic time, 

1 — oc a , 

or the square of the periodic time varies as the cube of the 
mean distance. 

375. Schol. The quantity \i which enters into the results of 
the preceding investigations is the value of the accelerating 
force at the unit of distance from the center of force. Nov/ 
the attractive forces of the sun and planets vary directly as 
their masses, and if M be the number of units of mass of the 
sun, and m the same of any planet, and if we assume for the 
unit of force the attraction of a unit of mass at a unit of dis- 
tance, M will express the attractive force of the sun at the unit 
of distance, and m that of any planet ; and the whole force by 
which they will tend to approach each other, or the whole force 
which the sun, regarded as fixed, exerts on the planet at the 
unit of distance is M.+m=[i ; and for any other distance r, 

r 2 r 2 ' 

The intensity of the solar and planetary attractions may be 
expressed in terms of terrestrial gravity. For this, let r, be 
the radius of the earth, and m l its mass, and since the forces 
are directly as the masses, and inversely as the squares of the 
distances, we have 

in, M_ Mr| 
rf : ~7~ g : my g ' 
the fourth term being the attractive force of the sun at any dis- 

7)XT" 

tance r. Similarly, the attractive force of the planet is \g; 

, • • . • n • M+m r\ u, r\ . .' 

hence their lomt influence is — - — . — fi"=-j. — g 9 in which 



CENTRAL FORCES. 235 

r 2 
— L .o- is that which was assumed above for the unit of force, or 

the attraction of the unit of mass at the unit of distance. 

376. EXAMPLES. 

Ex. 1. Required the velocity and time when a material 
particle is attracted to a fixed point by a force varying in- 
versely as the square root of the distance. 

ii il 

Ans. v=2i-r(a 2 —x 2 ) 2 

2 



i i 

2 



Ex. 2. Find the velocity and periodic time of a body re- 
volving in a circle at a distance of n radii from the earth's 
center. 

Ans. v= ( — 
V n 

V g 

Ex. 3. Find the least velocity with which a body must be 
projected from the moon in the direction of a line joining the 
center of the earth and moon, so that it may reach the earth. 

Ex. 4. Given the velocity, distance, and direction of pro- 
jection, when the force is attractive and varies as the distance, 
to find the orbit. 

Ex. 5. A body is projected in a direction which makes an 
angle of 60° with the distance, with a velocity which is to the 
velocity from infinity as 1 : VS : the force varying inversely 
as the square of the distance. Find the major-axis, the posi- 
tion of the apsis, the eccentricity, and the periodic time. 

Ex. 6. A body is projected at a given distance from the 
center of force with a given velocity, and in a direction per- 
pendicular to the distance, when the force is repulsive and 
varies inversely as the cube of the distance. Find the path of 
the body. 



HYDBOSTATICS. 



CHAPTER I. 

377. Matter exists in the various states of solid, fluid, and 
aeriform. In solid bodies the homogeneous integrant particles 
cohere firmly, and do not admit of interchange of position. 
The homogeneous integrant constituents of fluids possess less 
cohesion, and change their position among each other on the 
application of a moderate force. 

Fluids differ from each other in the degree of cohesion of 
their constituent parts and the facility with which they will 
yield to the action of an impressed force. 

378. A perfect fluid is one in which the cohesion of the con- 
stituent particles is so feeble that the least force is capable of 
effecting a separation and causing motion among each other 
in all directions. Perfect fluids are the only ones which will 
be made the subject of investigation. 

Fluids are divided into two classes, incompressible or liquid, 
and compressible or aeriform. 

379. Incompressible fluids are those which retain the same 
volume under a variable pressure. They may be made to take 
an infinite variety in form, which form they will retain when 
acted upon by no external pressure nor accelerating force. 
Of this class water may be regarded as the type. 

380. Compressible fluids, of which atmospheric air is the 
type, change at once in form or volume with every variation 
of pressure which they sustain, and return again to the same 
form and volume when the same circumstances of pressure and 
temperature are restored. 



EQUALITY OF PRESSURE. 



237 



PRINCIPLE OF EQUALITY OF PRESSURE. 

381. Prop. A force or pressure applied to a given portion of 
the surface of a fluid is transmitted, undiminished in intensity, 
to every other equal portion of the surface. 

A force impressed on a solid is effective only in the direc- 
tion of its action, and is sustained by an equal force impressed 
in an opposite direction. Applied to a fluid the force is effect- 
ive in every direction, and can be sustained only by forces ap- 
plied to every point in the surface of the fluid. 

Let the annexed figure 
represent a section of a 
vessel filled with a fluid 
without weight. In this 
case, supposing the press- 
ure of the atmosphere re- 
moved also, the surface of 
the fluid would be subject 
to no pressure, and would 
retain its form if the sides of 
the vessel were removed. 

Let an aperture of a giv- 
en size, as one square inch, 
be made in either side, AB, 
and a piston be accurately fitted to it, and supposed to move 
without friction. Now, since the particles are without friction, 
a force Pj of one pound, applied to this piston, will act upon 
the stratum of particles in contact with its base, and this stra- 
tum upon the next, and so on. The force will therefore be 
transmitted to the opposite side, EF, and will cause motion in 
the equal piston P', , unless counterpoised by an opposite force 
of one pound. If this counterpoise be applied, then, since the 
particles move freely among each other without resistance 
from friction or cohesion, the particles in the direction of the 
applied forces will act on those lying without, and will commu- 
nicate motion to them, unless resisted. A motion will therefore 
be communicated to any equal piston P',', unless sustained by a 




238 HYDROSTATICS. 

pressure of one pound. Hence both pistons V\ and P',' will 
sustain the same pressure by the action of P,. 

Further, if the pistons P' a and P',' were contiguous, or one 
of them twice the size of P 1? then it is obvious that a force of 
two pounds would be necessary to preserve the equilibrium ; 
or, if the base of the piston P 4 be four times that of P x , then a 
force of four pounds would be necessary to keep P 4 at rest; 
and if the base of one piston be n times that of the other, the 
pressure on the former will be n times that on the latter, or a 
force of one pound will produce a pressure of n pounds, where 
n may be the ratio less one, of the whole surface of the vessel 
to the base of the piston. 

Cor. 1. The pressures P a and P n on any two portions A. x 
and A„ of the surface will be proportional to their areas, or 

hk= n - (134) 

Also, the normal pressure p on a unit of surface will be 

Cor. 2. Every stratum of particles in the interior, of the 
same dimensions as the base of the piston, wherever situated 
and however inclined, is subject to a pressure equal to that ap- 
plied to the piston ; and since the fluid is without weight, every 
particle presses and is pressed equally in every direction. 

Cor. 3. If all the pistons except P and P' be firmly fixed, 
and P be forced in, since the fluid is incompressible, whatever 
fluid is displaced by P will be forced into the tube P'. If h and 
h' be the spaces through which the pistons move, a and a 1 the 
areas of their bases, then 

ha=h'a'. 
But P and P' being the equilibrating pressures on these areas, 

«_P_ 
a'~P /5 
hence hV=h'V', or hY-h'V'=0, 

or the principle of virtual velocities applies to fluids in equilib- 
rium. 




SURFACES OF EQUILIBRIUM. 239 



SURFACES OF EQUILIBRIUM. 

382. Prop. The free surf ace of a fluid subjected to the action 
of an accelerating force is, when in equilibrium, perpendicular 
to the direction of the force. 

A liquid void of gravity will take and retain any form im- 
pressed upon it ; but when subject to the action of an acceler- 
ating force, a containing vessel will be necessary in order to 
preserve a coherent mass. 

Let ABCD be the section of a ves- 
sel containing a fluid subject to the ac- 
tion of gravity, the base DC being hori- 
zontal. If any portion of the free sur- 
face, as abc, have a direction not per- 
pendicular to the direction of gravity, 
this force may be resolved into two 
components, one of which is parallel to this surface ; and since 
the particles are free to move in that direction as down an in- 
clined plane, they will yield to this component. But whenever 
every portion of the surface is perpendicular to the direction 
of gravity this force will have no component in the direction 
of the surface, and every portion of it will be urged vertically 
downward with equal intensity. 

Cor. Hence the surfaces of fluids at rest, and acted on only 
by gravity, are horizontal. But since the directions of gravity, 
acting on particles remote from each other, are convergent to 
the center of the earth nearly, the surfaces of large masses of 
fluid are not plane, but curved, and conform to the general 
figure of the earth. 

383. Prop. If a vessel containing a fluid be moved horizontal- 
ly with a constant accelerating force, the surface will take the 
position of an inclined plane. 

Let the vessel ABCD containing a fluid be moved horizon- 
tally with a uniformly accelerating force P. Then any ele- 

w 
ment, k, of the surface, whose weight is w, and mass - (22), 

will be urged by its weight w in the direction kg, and by its 



240 



HYDROSTATICS. 



W 

inertia P— in the direction kh, and 

g 
therefore by their resultant in the 
direction ko. As the same is true 
of every element of the surface, the 
whole surface (Art. 382) will be per- 
pendicular to ko. 
Let the angle of inclination to the horizon HIM=I. Then, 




since JIIM.=okg, 



w 



tan. okg—t&n. I=— —=— . 



w g 

384. Prop. If a vessel containing a fluid be made to revolve 
uniformly about a vertical axis, the surface of the fluid will take 
the form of a paraboloid of revolution. 

Let the vessel ABCD revolve uni- 
formly about the vertical axis XX x . 
Any element k of the surface will be 
urged horizontally by a centrifugal force 
directed from the axis, and vertically 
downward by gravity. Let o be the 
angular velocity of the fluid, w the weight 
Xi of the element k, and y its distance kM 

from the axis. Then (82) we have, for the centrifugal force P 
of this element, 




P=< 



w 



•y 



g 



Being urged vertically downward by its weight w, the surface 
of the element will be perpendicular to the resultant &R=R 
of these two forces. If this resultant be produced to meet the 



axis in N, we have 



and 



or 



Rg kM 



kg 

, w 
cry— 

—I: 
W 



MN' 



y 

'MN* 




PRESSURES ON IMMERSED SURFACES. 241 

.-. the subnormal MN=-= a constant, a property of the 
parabola. Hence the surface is a paraboloid of revolution. 



NORMAL PRESSURES ON IMMERSED SURFACES. 

385. Prop. The pressure on the horizontal base of a vessel 
containing an incompressible fluid is proportional to its depth 
below the surface of the fluid, and independent of the form of the 
vessel. 

Let HO be the free surface of the fluid in 
the vessel ABCD, whose base DC is horizon- x \ / 

tal, and suppose the fluid divided into hori- 
zontal strata of small but equal thickness. 

Let a l9 a 2 ,a 3 , &c, denote the successive 
strata, 

A , A„ A 2 , the units of surface in each, 

P , Pj, P 2 , the whole pressure on each, 

p , p l9 p 2 , the pressure on a unit of each, 
a the common thickness of the strata, and o the density of the 
fluid. Since the thickness of each stratum is supposed indef- 
initely small, the upper and lower surfaces of each may be re- 
garded as equal. 

Now the weight of a x is w 1 =gpaA (24), 
" " " " a 2 is w 2 =gpaA lt 

" " " " a n is w n =gpaA n _ 1 . 
The pressure on a x is P , 

P 

on a unit of a 1 is -7-^=^0 ( 13 5), 

on a 2 is P 1 =P +^ 1 =P +gpaA , 

.P. P ft gpaA a 
on a unit of a 2 is — -=— H — - — or p , —p +gpa, 

on a 3 is Y 2 .=l ) 1 +w 2 ='P l +gpaA l9 

..." . P 2 P. gpaA, * 
on a unit of a z is J^=J^+-~^ — or p 2 =p 1 +gp.a=p +gp.2a f 



on a n+l is 'P n =V n _ l +w n =V n _ l +gpaA n _ l9 

Q 



242 HYDROSTATICS. 

. ,- . P„ P n _! gpaA n _, 
on a unit of a n+l is — =— + br A or p n =p +gp.na. 

But na=h is the depth of a n+1 below the surface of the fluid, 
and if the upper surface of a n+l represent the surface of the 
base, A the number of units in the base, then the pressure p on 
a unit of the base will be 

P-P*+gph, 
and the whole pressure on the base 

~P=Ap = Ap +gphA=~P +gphAarh. (136) 

Cor. 1. Since h A is equal to a prism whose base is A and 
height h, and gphA is its weight, if we disregard P , which 
may represent the pressure of the atmosphere on the surface 
of the fluid, we have, for the pressure of the fluid on the base 
of the vessel, the weight of a column of the fluid whose base is 
that of the vessel, and height the height of the surface of the fluid 
above the base. It is obviously immaterial whether the surface 
pressed is that of the base of the vessel or a horizontal surface 
of an immersed solid. 

Cor. 2. Since a cubic foot of water weighs 1000 oz.=62.5 
lbs., we have, for the pressure on the base of any vessel con- 
taining water, 

P=62.5 hk lbs., (137) 

where h is the height in feet of the surface of the water above 
the base, and A the number of square feet in the base. 

Cor. 3. The pressure on every portion of a horizontal stra- 
tum of the fluid will be the same, and since this pressure is 
transmitted equally in every direction, the pressure on every 
element of the sides of the vessel having the same depth will 
be equal to that on the surface of the stratum. If the sides of 
the vessel are inclined, this pressure will be the normal press- 
ure on the sides at that depth. Let the annexed figures, A, 
B, C, D, represent vessels of different forms and capacities 
with equal bases. Then the pressures on the base of each 
will be equal when filled to a common height, and the points 
a, b, c, e, g, h, k, I, m, having a common depth, will be equally 
pressed normally to the surfaces. In B the horizontal surface 
df of the vessel will experience the same pressure vertically 



PRESSURES ON IMMERSED SURFACES. 243 

A B C D 



upward as that to which the stratum^ of the fluid is subject. 
Moreover, the stratum fg- being in equilibrium, must be press- 
ed equally upward and downward. The downward pressure 
is equal to the area of the section fg into the depth h of the 
section below the surface of the fluid in the vertical branch. 
But the upward pressure is due to the pressure of the fluid in 
the inclined branch ; and since the area of the section fg is 
common to the measures of both pressures, the vertical heights 
of the fluid in both branches are necessarily equal. 

386. Prop. The normal pressure on any plane surface in- 
clined to the horizon is proportional to the depth of its center 
of gravity below the surface of the fluid. 

Let k be any indefinitely small element of the immersed sur- 
face, x its depth, and p the pressure it sustains. By the pre- 
ceding proposition, 

p—gpxk; 

and since the pressure is the same in every direction, p will be 
the pressure normal to this element, whatever be its position 
or inclination. The expression for the normal pressure on 
every other element will be of the same form. Hence the 
whole normal pressure P will be 

P= 2gpxk=gp'Exk. 

Let x be the depth of the center of gravity of the immersed 
surface below the surface of the fluid, and A its area. Then 
(29), Ax = 2x k; 

... ~P=gpAxccx. (138) 

Cor. Since gpAx is the weight of a prism of the fluid whose 
base is A and altitude x, the normal pressure on any immersed 
surface inclined to the horizon, or on any side of the contain- 
ing vessel, is equal to the weight of a fluid prism whose base 



244 



HYDROSTATICS. 




is the surface pressed, and height the depth of its center of 
gravity below the fluid surface, or, if the fluid be water, 
P=62.5 A^ lbs. (139) 

387. Prop. To find the pressure on any immersed surface or 
side of a vessel in a given direction. 

pj j3 Let P be the normal pressure on 

the inclined side, CDFE=A of the 
vessel CBE, and let it be required 
to determine the horizontal and 
vertical pressures on A. Let a be 
the angle which the side A makes 
with a vertical plane, and |3 its in- 
clination to the horizon. Then, 

since PAPj=a and PAP 2 =ft resolving P horizontally and 

vertically, we have 

P,=P cos. a=gpxA cos. a, 

and P 2 =P cos. p=gpxA cos. 13. 

But CDHG=A a is the projection of the side A on a vertical 

plane, and EFHG=A 2 is the projection of the same side on a 

horizontal plane. 

Therefore, A,=A cos. a and A 2 =A cos. ft 
and V l =gpA 1 x and ~P 2 =gpA 2 x. 

Hence the horizontal pressure on A is equal to the weight 
of a prismatic volume of the fluid whose base is the vertical 
projection of A and height the depth of its center of gravity ; 
and vertical pressure, a volume whose base is the horizontal 
projection, and height the depth of the center of gravity of A. 
And generally the pressure on any immersed surface in a given 
direction is equal to the weight of a prismatic column of the 
fluid whose base is the projection of the surface on a plane per- 
pendicular to the given direction, and height the depth of the 
center of gravity of the surface below the surface of the fluid. 

388. Prop. To find the resultant pressure of a fluid on the in- 
terior surface of the containing vessel. 

Let ABCD be a section of a vessel of any form filled with 
a fluid. When the sides of the vessel are curved, the normal 



PRESSURES ON IMMERSED SURFACES. 



245 




pressures on the elements of the 
surface will not be parallel. But 
we may resolve the normal press- 
ure on each element into vertical 
and horizontal components, and 
each horizontal component into 
two others parallel to two rectan- 
gular axes. We shall then have 
to find the resultants of three sets 
of parallel forces, and, finally, the resultant of these three re- 
sultants. We may, however, determine the resultant more 
simply, as follows : 

1°. The resultant of the horizontal pressures. 

Let HK be the intersection of a vertical plane by the plane 
ABCD produced, and let k be the vertical projection of any 
element k x on this plane. Now the normal pressure on the 
element k x of depth a 1 k 1 =h is M,, and the horizontal press- 
ure (Art. 387) is hk. The horizontal pressure on each ele- 
ment being of the same form, the whole horizontal pressure 
perpendicular to HK of the surface convex toward HK will 
be 2.M. But to every element k x of the surface convex to- 
ward HK there is a corresponding element k 2 of the surface 
concave toward HK, the projection k of which is the same. 
Hence the horizontal pressure on k 2 is —hk, and the pressure 
on the whole surface concave toward HK is —Zhk. The re- 
sultant, therefore, of all the horizontal pressures perpendicular 
to KH is 2.M— 2.M=0. Since the same is true on whatever 
side of the vessel the plane HK be drawn, the resultant of all 
the horizontal pressures will be zero. 

2°. The resultant of the vertical pressures. 

The pressure on the element k Y vertically downward is the 
weight of a filament of the fluid whose base is k x and height 
a l k l . The pressure on h! vertically upward is equal to the 
weight of a filament of the fluid whose base is k' and height 
a x k'. The resultant of the pressures on the corresponding ele- 
ments k' and k x is therefore equal to the weight of the fila- 
ment of fluid k'k x . In the same manner, the resultant of the 
vertical pressures on any two corresponding elements will be 



246 HYDROSTATICS. 

the weight of the filament of fluid whose bases are these ele- 
ments. The resultant of all the vertical pressures will then 
obviously be the sum of the weights of these filaments, or the 
weight of the fluid. 

Hence the resultant of all the partial pressures on the in- 
terior surface of the vessel is equal to the weight of the fluid, 
directed vertically downward, and will pass through the cen- 
ter of all the parallel vertical pressures {Art. 44), or the center 
of gravity of the fluid {Art. 94). 

Cor. Since the horizontal pressures balance, there will be 
no resultant pressure to cause motion in 
the vessel horizontally. If, however, 
an aperture be made on one side of the 
vessel, the pressure on this portion of 
the vessel will be removed, and the 
pressure on the corresponding opposite 
portion will tend to produce motion in 

that direction, and if the vessel be free to move, motion will 

actually ensue. 

389. Prop. To find the resultant pressure on any solid im- 
mersed in afiuid, and its point of application. 

1°. The horizontal pressure on any element of the immersed 

solid is the same as that on the 





H , ,*S 

vertical projection of the element 

{Art. 387), and the entire press- 
ure on each of the opposite sur- 
faces of the solid will be equal to 
that on their vertical projections. 
But the projections of any two op- 
posite surfaces on the same vertical plane are equal. Hence 
the horizontal pressures on any two opposite surfaces are equal, 
and the resultant of the horizontal components of the pressures 
in every direction is zero. 

2°. The vertical downward pressure p' on any element h! 
of the upper surface is equal to the weight of a filament of the 
fluid ak! whose base is k' and altitude ak'=h', or if k be the 
horizontal transverse section of the filament through the cen- 



PRESSURES ON IMMERSED SURFACES. 247 

ter of gravity of k\ p'=gph'k. But the sum of all the filaments 
resting on the upper surface of the solid is equal to the volume 
V' of the fluid vertically above the solid, and the whole down- 
ward pressure P' is the weight of this volume, or 
?'=2.p'=gp2.h , k=gpV'. 
In the same manner, the vertical upward pressure p 1 on the 
corresponding element k x of the lower surface is equal to the 
weight' of a filament ak l of fluid whose base is k x and height 
ak 1 =h li or p 1 =gph 1 k. But the sum of all these filaments is 
the volume V, of the solid and fluid above the upper surface, 
and the whole upward pressure P x is the weight of this volume 
of fluid, or 

Now h 1 —h'=h is the length of the filament k'k x of the solid, 
p 1 —p'=p its weight, and V,- V'=V is the volume of the 
solid. 

Therefore, the difference P of the upward and downward 
pressure is 

and 1 P=gp2.h l k—gp'2.h'k=gp2hk=gpV ; 

... p =g . p v=2!p, (140) 

or the resultant pressure is vertically upward, and equal to the 
weight of a mass of the fluid of the same volume as the solid. 
3°. To find the point of application of this resultant. Let 
the distance of the point of application of p=hk from the ver- 
tical plane HK be x, and the distance of the point of applica- 
tion of the resultant P be x. Then, since the moment of the 
resultant equals the sum of the moments of the components, 
Vx=^.px, 

and z=-^-=^--. (141) 

But (29) the point thus determined is the center of gravity 
of the displaced fluid. 

Hence the resultant of all the pressures on the immersed 
solid is equal to the weight of the displaced fluid, acts vertical- 
ly upward, and its point of application is the center of gravity 
of the displaced fluid. 



248 



HYDROSTATICS. 



390. Prop. To find the conditions of equilibrium of an im- 
mersed solid. 

The conditions of equilibrium involve a consideration of the 
weight of the body. Let V be its volume, and a its density. 
Its weight will be gaV. The body will be urged downward 
by a force equal to gaV applied at its center of gravity, and 
by Art. 389 it is urged upward by a force equal to gpV, since 
the volumes of the solid and displaced fluid are equal. 

In order to equilibrium, then, 1°. These forces must be equal; 
and, 2°. Their lines of direction coincident. 

The first condition gives 

g aV=gpV f 
or o=p ; 

that is, their densities must be the same. 

Cor. 1. If a be not equal to p, the body will ascend or de- 
scend by a force equal to g(p— o)V, according as p—o is posi- 
tive or negative. If p>a, the body will rise to the surface, and 
be but partially submerged. Let V x be the displaced fluid, or 
the part of the solid immersed when the equilibrium is restored. 
Then 

gpV^goV, (142) , 

and V : V,=p : o; 

or the whole volume of the solid is to the part immersed as the 
density of the fluid is to the density of the solid. 

Cor. 2. If the centers of gravity of the solid and displaced 
fluid be not in the same vertical line, the body will be acted 
upon by two parallel forces in opposite directions, and will 
cause the body to turn round. The point of application of the 
resultant of these forces may be found by Art. 29. 

391. Def. The section of a floating body made by a plane 
coincident with the surface of the fluid is called the plane of 
flotation. The line joining the centers of gravity of the solid 
and of the displaced fluid is called the axis of flotation. 

DEPTH OF FLOTATION. 

392. Prop. To find the depth of flotation when the volume 
and density of the body are known. 



DEPTH OF FLOTATION. 249 

Let V be the volume of the body, o its density, Y x the volume 
of the displaced fluid, and p the density of the fluid. Then (142) 
gaY=gpY 19 

and V^-V. 

9 

Now whenever Y x can be determined in terms of the depth 
of flotation x, this expression will suffice to determine x. 

If the solid be a right cylinder whose axis a is vertical, and 
the radius of whose base is r, we have 

Y 1 =nr 2 x, 

and Y=nr 2 a; 

a 
.'. x=-a. 
P 

If the body be a right cone with the axis vertical and vertex 

downward, let r be the radius of the base, and a the altitude. 

r 

Then the radius of the plane of flotation is -x. Hence 
r a 

r 2 

V =l7T— T 3 

and V=|7T?* 2 tf; 



x=a<J - 



P 
If the vertex of the cone be upward, 



v/-;. 



P 

393. Prop. To find the positions of equilibrium of a right 
triangular prism when the axis is horizontal. 

Whenever a body can be conceived to be generated by the 
motion of a plane surface perpendicular to itself, and the body 
floats with its axis horizontal, it is evident that, in whatever 
position it be turned, its center of gravity and that of the part 
immersed will lie in the same vertical section ; and, further, 
that the center of gravity of the body will be at the center of 
gravity of the section, and the center of gravity of the im- 
mersed part at the center of gravity of its section. Also, the 
ratio of the part immersed to the whole mass is the same as 



250 



HYDROSTATICS. 



the ratio of their sections. We can therefore limit the investi- 
gation to the positions of equilibrium of one of the generating 



sections. 




Let ABC, Fig. 1, be a 
generating section when 
only one angle is im- 
mersed, and Fig. 2 when 
two angles are immersed. 
LetAB=c,AC=6,BC=a, 
AQ=£, AP=y, and let a be 
the ratio of the density of 
the solid to that of the fluid. 



Now from (142) we have, in Fig. 1, 

AQP__±xy sin. A_xy 
ABC "~ Ibc sin. A ~~ be' 



o= 



In Fig. 2, 



o= 



.'. xy=o.bc. 
BQPC ABC-AQP bc-xy 



(a) 



ABC 



be 



0) 



ABC 

.*. xy=bc(l—o). 

Since (b) is derived directly from (a), by changing a into 
1— a, we may examine the case of Fig. 1, and deduce that of 
Fig. 2 from the result, by changing a into 1— a. 

Bisect BC in M, and QP in m. Join AM and Am. Take 
MM'=£MA and mm'—\mA. W and m' are the centers of 
gravity of ABC and AQP. Join Mm and Wm'. It is obvious 
that the center of gravity of BQPC is in the line M'wi' produced. 
Now the second condition of equilibrium requires that Wm' 
should be vertical, and since Mm is parallel to M'm', Mm is 
perpendicular to QP and MQ=MP. Let AM=h, MAQ=0, 
and MAP=</>. Then 

MQ, 2 =x*+h 2 -2hx cos. 6, 
and MF 2 =y 2 +h 2 -2hy cos. 0. 

Hence x 2 — y 2 — 2hxcos. 0+2hy cos. 0=0. (c) 

Substituting in (c) the value of y from (a), and reducing, 

x'-2h cos. 6.x 3 +2obch cos. (f>.x-o 2 b 2 c 2 =0, (d) 



DEPTH OF FLOTATION. 251 

and changing a into 1— o, we have for Fig. 2, 

x i -2hcos.6.x 2 +2(l-o)bchcos. 0.z-(l-(7)W=O. (e) 

The values of x deduced from (d) and (e), and the correspond- 
ing values of y from (a) and (b), will give the positions of equi- 
librium required. 

Now, since the degree of the equation is even and the abso- 
lute term is negative, there are at least two possible roots, one 
positive and the other negative. The other two roots may be 
real or imaginary. If real, Descartes' rule of signs indicates 
that three will be positive and one negative. Hence, since the 
values of x and y, which are applicable to the question, are 
necessarily positive, equations (d) and (e) indicate no more 
than three positions of equilibrium. The values of x must also 
be less than b, and when subtituted in (a) and (b) give for y 
values less than c. 

Let us take the case of an equilateral triangle. 

Then d=<f>=30, a = b=c, and h—b cos. 6= c cos. 0. 

These values, substituted in (d), give 

x* - \ax* + %ca % x - a 2 a' = 0, 
or (x'-oa 2 ) (x 2 -%ax+oa 2 )^i). (J) 

This equation is satisfied by putting 
x 2 — aa 2 =0; 
whence x =aV<?, 

the negative value of x being inapplicable. 

Since tf<l, we have x<^a, and the value of x indicates an 
actual position of equilibrium. But from (a) we have 

o.a 2 a. a 2 — 

y= = — —=a va=x, 

x a\fo 

or one side of the triangle is parallel to the surface of the fluid. 

Equation (/) is also satisfied by making 

x 2 -Ux+oa 2 =0; 



whence x =-{3±V9—16o}. (g) 

Now, in order that these values of x may be real, we must 
have 

16a<9 or 16(7=9. 



252 HYDROSTATICS. 

9 9 

Hence <r<— or o=— , 

or o can not be greater than — . 

But if V9— 16<7>1, then x>a, which is inconsistent with the 
supposition that only one angle is immersed. The greatest 
value, therefore, which V9—l6o can have is unity, and in this 
case the second angle will lie in the surface of the fluid. 

Putting V9-lQo=l, 

we get o=-, 

for the least value a can have when one angle only is im- 
mersed. The limits of a for this case are, therefore, 

1 A 9 

2 and if 

If in (g) we cnange o into 1— o, we get 

x=^\3±V9-lQ(\-o)\ 



=-\3±Vl6o-7\. (h) 

From which it may be shown that the limits of the value of 
o for this case are 

7 a l 
16 and 2* 

Since equations (d) and (e) may have three real positive 

roots each, there may be three positions of equilibrium for 

every single angle immersed, and three for every two angles 

immersed, and therefore eighteen in the whole. But in the 

particular form considered above these are not all possible, 

except *=«, in which case x=a or y=a; that is, either the 

angle B or C lies in the surface of the fluid. This would ren- 
der six of the positions pertaining to Fig. 1 the same as six of 
those pertaining to Fig. 2, making but twelve really different 

1 9 

ones. If ff>- and <7<t^, there will be nine positions with 

ii It) 



CENTER OF PRESSURE. 253 

one angle immersed and three with two immersed ; if <j<- 

7 7 

and <t>tt:» n ine of the latter and three of the former ; if a<— 
16 lb 

g 

or o>—, there will be only three of each. 
lb 



CENTER OF PRESSURE. 

394. Def. The center of pressure in any immersed surface 
is the point of application of the resultant of all the pressures 
upon it. It is therefore that point in an immersed surface or 
side of a vessel containing a fluid, to which, if a force equal 
and opposite to the resultant of all the pressures upon it be 
applied, this force would keep the surface at rest. 

395. Prop. To find the center of pressure on any immersed 
plane surface. 

Let the immersed surface 
be the inclined side AE of 
the vessel ABE supposed to 
be filled with a fluid. Let x 
and y be the co-ordinates of 
any element k referred to 
the rectangular axes AY and 
AX, and hk=h the vertical 
distance of the element be- 
low the surface of the fluid. 
Then the normal pressure p on k will be 

p=gphk. 

The pressure on each element will be of the same form, and 
their sum or resultant R will be 

R=2.p. 

The moment of each partial pressure, in reference to the 
plane through AY perpendicular to AX (Art. 46), will be px, 
and the moment of each, in reference to the plane through AX 
perpendicular to AY, will be py. Hence, if x and y are the 




254 HYDROSTATICS. 

co-ordinates of the point of application C of the resultant R, 
we have (Art. 43) 

R = 2.p, Rx=2px, Ry=2.py. 

If 6 be the inclination of the side of the vessel, or immersed 
surface to the surface of the fluid, we have 

h=x sin. 6. 
Hence p=gpxk sin. 6, 

and R=2.gpxk sin. 0=gp s'm. O.l.xk, 

Rx=2.gpx' 2 k sin. 0=gp sin. 0.2.x 2 k, 
Ry=2.gpyxk sin. Q=gp sin. O.^.xyh 
_ 2.x*k , x 

•*• x =i~x-r < 143 > 

_ Z.xyk 

and *=& < 144 > 

When the upper boundary is below the surface of the fluid 
and at a distance a from it, then, since the pressures p are 
limited to the immersed surface, h=(a+x) sin. 0, and we shall 
have 

_ S(ax+x*)k 



2(a+x)k ' 



(145) 



_ 2(a+x)yk 

rfefr < 146 > 

Further, if the axis of x is so taken that it will bisect every 
horizontal line of the immersed surface, the pressures on oppo- 
site sides of this axis will obviously be equal, and the center 
of pressure will lie in this axis or y=0. 

It will be observed that the numerator of (143) is the mo- 
ment of inertia of the surface, and that the denominator is the 
statical moment. Hence the denominator is equal to the area 
of the surface multiplied by the depth of its center of gravity 
(29). Hence, if A be the area and x^ this depth, 

*=*V < 147 > 

_ ^.xyk 

v=-xt- < 148 > 

396. Prop. To find the center of pressure of a rectangular 



CENTER OF PRESSURE. 



255 



surface vertically immersed, and having one side in the surface 
of the fluid. 

Let YYj be the line of intersection 
of the immersed surface ABCD with Ti- 
the surface of the fluid, and conceive 
ABCD divided into n rectangles by- 
horizontal lines, n being a very large 
number. Let AB=6 and AD=h, 
and draw OX bisecting the rectangle. 
The center of pressure is obviously 
in OX and y—0. To find x, we have, since the height of each 

small rectangle is -, for the area of each, 




, bh 
n 



The distances of these rectangles from the surface are 



their squares, 
Hence 



n 
bh 



h 2h 3h e 

-, — , — , &c. ; 
n n n 

If 2Vf 37i 2 s 

n n n 

bh 3 bh 3 .2 2 bh\& e 



=— (l' + 2 9 +3 9 + n*). 



n 



But when n is very large, the sum of the mth powers of the 
natural numbers 1,2, 3, &c, to n is 



wi+1 



Also, 



„ bh 3 n 3 
\ 2.x*k=- T .-=m 3 . 
n 3 3 

Ax y =bh.\h=\bh\ 

• T - * =2-h 

- x ~m* 3 ' 



or the center of pressure is two thirds the height of the rect- 
angle below the surface of the fluid. 

397. Prop. To find the center of pressure when the immersed 



256 



HYDROSTATICS. 



surface is a triangle, having one side horizontal and the opposite 
vertex in the surface. 

Let BDC be the triangle, h its height, 
and b its base, and suppose the triangle di- 
vided as before into n horizontal divisions. 
The heights of the successive divisions 
h 



A F B 

M 

dI£ — i ic 



E 

the lengths, 
the depths, 



are - ; 
n 

b 2b 36 



n n 



, &c. ; 



_h _2h _Sh 
X >~n' X2 ~^' X3 ~~n~' 



Their areas, considering the horizontal sides of the success- 
ive trapezoids as differing insensibly from each other, are 

— hh h —— h - Sbh 

Multiplying these by the squares of their depths, and taking 
their sum, we have 



Q7 bh* 2 3 M 3 3W c 
2.x*k=— +^-+-^+, &c. 



n*bh 3 



n* 



bh" 



=-^-(l 3 +2 3 +3 3 +, &c O, 

n 4 * 4 4 



Also, 



Hence 



K=\bh, and x 1 —%h. 
A^, = ^bh\ 



\bh* 



#>h % 



4h. 



If now BE be drawn bisecting the base, the center of press- 
ure will be in BE, and at a distance from the surface BO=fBE. 

398. Prop. To find the center of pressure when the base of the 
triangle lies in the surface. 

In this case the heights of the successive divisions of the tri- 
angle ADB {Fig., Art. 397) are - ; 



2b , Sb 



the lengths, b — , b , b 



, &c. ; 



CENTER OF PRESSURE. 



257 



u j u h 2h 3h * 

the depths, x,=-, x„ = — , x= — , ofcc.; 

7 bh bh i bh 2bh bh Sbh • 

the areas, k= =-, k = r-, k s — — s &c, 

1 n n 2 " n n 2 3 n n 2 ■ 

which, multiplied by the squares of the depths, and their sum 
taken, give 

■-, M 3 6/* 3 2 2 Z^ 3 2 3 bh 3 3W S 5 bh 3 „ 

2.z 2 A:=— r H ; ; — 1 i 1 — h, &c, 

7l 3 ?l 4 7l 3 7l 4 7Z 3 7Z 4 

=- r (l a +2 , +3'+ . . . n 2 ) r (l 3 +2 3 +3 3 + . . . n 3 ), 



bJ^T^ 






=±bh 3 -±bh 3 =^bh 3 , 



and 



A.Xi=\bh.\h=±bh 2 . 



Draw DF bisecting AB, and take FO' = |FD. O' is the cen- 
ter of pressure of DAB. 

399. Prop. To find the center of pressure when a rectangle is 
immersed vertically, having a side parallel to the surface of tlie 
fluid and at a given distance below it. 

Let a be the distance of EF, the upper 

side of EFCD from AB the surface of the 

fluid, and conceive the rectangle divided 

as before into n divisions whose heights 

h . bh TT . 

are — , and areas — . Using (145), and 
n n o \ / 

reckoning x from EF, the origin of the 
surfaces, the depths are 



M 










Hence 2.(a-\-x)xk= — r H — r + 

v ' n n n 



h 2h Bh 

a-\ — , a-\ , aH , &c. 

n n n 

abh 2 bh 3 2abh 2 2 2 bh 3 Sabh 2 S 2 bh 2 



-h 



-+- 



n 2 + ~ n 3 



■+,&c. 



abh 2 , x bh 3 , m m o 

=— -(1 +2+3+ . . . . n) +— (l 2 +2 2 +3 2 + . . . O 



abh'n 2 bh 3 n^ 
? 3 
R 



= ^-2 + ^3 = ^ 2+ ^ 3 - 



258 



HYDROSTATICS. 



And l(a-{-x)k=Ax 1 



bh(a+\h)=abh+\bh\ 
\abtf-\-\bh ? _ i 2h+Sa 
abh+Uh 2 



Hence 
in which d 



hO=a+x=\h 



2h+Sa 



■+a=: 



3 * h+2a 
d 3 -a* 



MO. 




h+2a ' ~ 3 'd*-a*' 
AD=a+h. 

400. Prop. To determine the conditions of equilibrium of 
fluids of different density. 

B When fluids of different densities which 
do not mix or unite chemically are contain- 
ed in the same vessel they will arrange them- 
selves in the order of their densities, the most 
dense taking the lowest position. This is ob- 
vious from the ready displacement of their 
particles and Art. 390, Cor. 1. 

The limiting surfaces of each will likewise be horizontal ; 
for, in order to equilibrium, any horizontal stratum must sus- 
tain in every part of it the same pressure. But if the common 
surface, as EF, of two of them be inclined, then a horizontal 
stratum, as HR, will sustain at different points, as O and O', 
superincumbent columns of fluid of different weights, and the 
equilibrium will not subsist. 

401. Prop. In communicating tubes, the 
heights at which fluids of different densities 
will stand above their common base when in 
equilibrium, are inversely as their densities. 
In the bent tube AED, let one fluid oc- 
cupy the portion AR, and the other the por- 
tion RED, the fluids having the common 
base HR. Let h be the height of the fluid 
in AR above HR, and p its density, h' the 
height of the surface CD above HR, p' its 
density, and let the common base HR=A. 
The pressure of the fluid in AR on A is 
gphA, and that of the fluid in RED on the 
same base gp'h'A. In equilibrium these 
pressures will be equal. 



H 



B 



— D 



R 



EXAMPLES. 259 

Therefore gphA=gp'h'A, 

h p' 
and ^=j. (149) 

402. EXAMPLES. 

Ex. 1. A cubical vessel, each side of which is ten feet square, 
is filled with water, and a tube thirty-two feet long is fitted to 
an aperture in it, whose area is one square inch. If the tube 
be vertical, of the same diameter as the aperture, and filled 
with water, what is the pressure on the interior surface of the 
vessel, neglecting the weight of the water it contains ? 

Since the weight of a cubic foot of water is 1000 oz., the 

weight of one cubic inch is t^^= .5787 oz., and the weight 

of 32X12=384 cubic inches is 384X. 5787=222.2208 oz. = 

13.8888 lbs. This is the pressure on the aperture, or one 
square inch of the surface of the water in the vessel. 

The number of square inches in the interior surface of the 
vessel is 6X10 2 X 144=86,400. 

Then, Art. 381, the pressures being in the ratio of the areas 
pressed, the whole pressure on the interior surface will be 
P=13.8888X86399=1,200,000 lbs. nearly. 

Ex. 2. What is the pressure on the bottom of the vessel in 
Ex. 1 when the weight of the water in the vessel is taken into 
account ; 1°, without the vertical tube, and 2°, with it. 

1°. The base of the vessel is 100 square feet, and its depth 
10 feet. Hence, Art 385, 

P, = 10X100X62.5=62,500 lbs.=the weight of the water. 

2°. The pressure transmitted to each square inch of the base 
by the water in the tube is 13.8888 lbs. Hence the whole 
transmitted pressure P 2 = 13.8888 X 10 2 X 144 = 200,000 lbs. 
nearly, and the entire pressure on the base is P= 262,500 lbs. 

Ex. 3. What is the pressure on each vertical side of the ves- 
sel without the tube ? 

The depth of the center of gravity of each side of the vessel 
is 5 feet. Hence, Art. 386, 

P 3 = 5X10 2 X62.5=31,250 lbs., 
equal one half the pressure on the base. 



260 HYDROSTATICS. 

Thus we have the whole pressure on the interior surface of 
the vessel, including that produced by the water in the tube, 

P+P 1 +4P 3 =a,200,000 
+ 62,500 
+ 125,000 
= 1,387,500 lbs. 
Ex. 4. Required the pressure on the base of a conical ves- 
sel filled with water, the radius of the base being r=5 feet, and 
the altitude a=\0 feet. 
By (137), P=62.5AA lbs., in which h=a, and A=7rr 2 . 

.-. P=62.57rr 2 tf=62.5X3.1416X25X 10=49,087.5 lbs. 

Ex. 5. Required the normal pressure on the concave surface 
of the cone of Ex. 4. 



The slant height of the cone is -\/a 2 +r 2 , and the concave 

vV+r 2 
surface is 2nr — . The vertical depth of the center of 

. 2a 
gravity ol the concave surface is — . 

o 

2a 



Hence (137), P 1 =62.5X--rrr vV+r 2 

o 

=62.5X fX 10X3.1416X5 ^125= 73, 175.1 lbs. 

Ex. 6. Required the vertical pressure on the concave sur- 
face of the same cone. 

Since the normal to the side of the cone makes the same an- 
gle with the axis of the cone that the side of the cone makes 
with the base, the cosine of the inclination of the normal to the 

. . r 

axis is 



Va 2 + r 2 
Hence the vertical pressure {Art. 387) 

P 2 =P, -=^==62.5X|a7rrV« 2 +r 2 .-^==62.5X|7rr 2 a, 

=|P (Ex. 4) = 32,725 lbs. 

Ex. 7. Required the resultant of all the pressures on the in- 
terior surface of the same cone. 

By Art. 388, the resultant of the horizontal pressures is zero, 



EXAMPLES. 261 

and the resultant of the vertical pressures is equal to the ex- 
cess of the downward pressure over the upward pressure. 

.-. R=P (Ex. 4)-P 2 (Ex. 6) = 16,362.5 lbs., 
and this is the weight of the water in the cone. 

Ex. 8. A rectangular parallelogram, whose sides a and h are 
26 feet and 14 feet respectively, is immersed in water with 
the side b in the surface, and is inclined to the surface at an 
angle <£=56°.35'. Required the pressures Y 1 and P 2 on the 
parts into which the parallelogram is divided by its diagonal. 

Ans. P x = 82286.5 lbs. 
P 2 = 164573.0 lbs. 

Ex. 9. When «=30 feet, 6=20 feet, and </>=59°.38', re- 
quired the pressures on the equal parts into which the paral- 
lelogram is divided by a line parallel to the horizon. 

Ans. P,= 121332| lbs. 
P 2 =363998 lbs. 

Ex. 10. When the parallelogram of Ex. 9 is vertical, how 
far from the surface must the dividing line be drawn that the 
pressures on the two parts may be equal ? 

Ans. #=21.213 feet. 

Ex. 11. A sphere 10 feet in diameter is filled with water. 
Required the entire pressure on the interior surface, and the 
weight of the water. 

Ex. 12. A vessel in the form of a paraboloid, with vertex 
downward, is filled with water, and revolves uniformly on its 
axis. Required the time of one revolution when the angular 
velocity is just sufficient to empty it. 

Ex. 13. The concave surface of a cylinder filled with fluid 
is divided by horizontal sections into n annuli in such a man- 
ner that the pressure on each annulus is equal to the pressure 
on the base. Given the radius of the cylinder. Required its 
height, and the breadth of the m\h annulus. 

Ex. 14. A regular tetrahedron, with one face horizontal, is 
filled with water. Compare the pressures on the base, and on 
the other sides with the weight of the water. 




262 HYDROSTATICS. 

Ex. 15. An iron vessel 40 feet long, every transverse sec- 
tion of which is an isosceles triangle whose base is 16 feet and 
altitude 20 feet, floats with its vertex downward. If a cubic 
foot of iron weighs 487.5 lbs., required the depth to which the 
vessel will sink when the sides and ends are one quarter of an 
inch thick and there is no deck. 

Ex. 16. An embankment of brick- work, 
of which ABC is a section, weighs 120 
lbs. to the cubic foot. Its height AB is 
14 feet, and its base BC is 6 feet. Find 
whether or not the embankment will be 
overthrown by the pressure of water on 
the surface AB. 
H"-D B Draw AN bisecting CB, and take 

NG=|NA. G will be the center of gravity of the section, 
and the weight W will act at G. Take AM=|AB, and M 
will be the center of pressure (Art. 396). The resultant P of 
the pressure of the water will act at M and will pass through 
G. The moment of P to turn the embankment over C will be 
P.CO, while the moment of W to resist the overturn will be 
W.CD. If P.CO>W.CD, the embankment will be overturned. 
Since the embankment is uniform throughout its length, as 
also the pressure on it, we may determine the stability by 
taking one foot in length. 

Now W=lXl4X6XiXl20=5040 lbs., 

and P=1X14X7X62.5=6125 lbs., 

OC = BM=iXl4 and CD = f X6=4. 
.♦. W.CD=20160, 
and P.CO=28583f 

The latter being greater than the former, the embankment 
will be overturned. 

Ex. 17. A wall of masonry, a section of which is a rectan- 
gle, is 10 feet high, 3 feet thick, and each cubic foot weighs 
100 lbs. Find the greatest height of water it will sustain 
without being overturned. 

Ex. 18. If the height of the wall be 8 feet, its thickness 6 
feet, and each cubic foot weigh 180 lbs., find whether it will 
stand or fall when the water is on a level with the top. 



CHAPTER II. 

SPECIFIC GRAVITY. 

403. Def. The specific gravity of a body is the ratio of the 
weight of the body to the weight of an equal volume of some 
other body taken as the standard of comparison, and whose 
specific gravity, therefore, is taken as the unit. 

Water is generally employed a*s the standard of comparison 
for solids and liquids, and atmospheric air for aeriform fluids. 

Cor. If v, p, o, and w be the volume, density, specific gravity, 
and weight respectively of one body, and v l9 p 1 ,o 19 w l the 
same of another body; then, since v=v x , 

o w g-P-v p , x 

— =— =-2Jl_ = il (150) 

<T l 10, g.p i V l p i 

or the ratio of the specific gravities of two bodies is equal to 
that of their densities. 

404. Prop. To find the specific gravity of a body more dense 
than water. 

If the body be immersed in water it will descend (Cor., Art. 
390). Let w be the absolute weight of the body, and w 1 its 
weight or the force with which it will descend when immersed 
in water. Then the loss of weight in water is w— w lt and this 
is equal to the upward pressure of the water, or to the weight 
of a volume of water equal to that of the solid. Hence 
w absolute weight 
~w—w x ~ loss of weight ' ^ '■_ 

The absolute weight as well as the loss of weight is ascer- 
tained by the hydrostatic balance, which differs from the or- 
dinary balance only in having a hook appended beneath one 
of the scale pans, to which the body may be suspended by a 
fine thread and allowed to sink in a vessel of water beneath it. 
The body is first placed in the scale pan and counterpoised by 



264 HYDROSTATICS. 

a weight w, and then suspended to the hook, and when im- 
mersed in the water its counterpoise w x again determined. 

405. Prop. To find the specific gravity of a body less dense 
than water. 

Since the body A is less dense than water, it will not de- 
scend in the water by its own gravity. Let a more dense 
body B be attached to it, and call the compound body C. 
Let w = absolute weight of A, 

W = " " " B, and w\= its weight in water, 

w"= " " " 0, and w'[ = " " « " 

Then w'—w\ =loss of weight of B, 

W " — W " x =z » « " « c, 

and (w" —w'() — (w' —w\)=\oss of weight of A, the upward 
pressure of the water on A, and therefore equal to the weight 
of a volume of water the same as that of A. 

But w"=w'+w. 

and, by substitution, 

{w"— w") — (w'— w' 1 )=w'+w— w'[— w'+w^w+w^— w'{. 

w 
.'. 0=-^-, Tr (152) 

Hence, add to the absolute weight of the body the difference 
of the weights of the more dense and compound bodies in 
water, and divide the absolute weight of the body by the sum. 

406. Prop. To find the specific gravity of a liquid. 

Let a body whose weight is w be weighed both in water 
and in the liquid, the weight in the former being w , , and in the 
latter w 2 . Then w — w 1 and w—w 2 are the weights of equal 
volumes of the two liquids. Hence 

w—w„ 

g= -; (153) 

w — w l v 

or, if an empty bottle whose weight is w, weighs when filled 
with water w 1 , and when filled with the liquid w 2 , then 

w 9 —w 

a=-^ . (154) 

w,—w v ' 



SPECIFIC GRAVITY. 265 

407. Prop. To find the weights of the constituents in a me- 
chanical composition when the specific gravities of the compound 
and constituents are known. 

Let w, Wj, w 2 be the weights of the compound and constitu- 
ents respectively. 

(7, a l9 o 2 their respective specific gravities, 
v, Wj, v 2 their volumes. 
In all merely mechanical combinations, 

v=v 1 +v 2 (a), and w=w , +w 2 (b). 

n . w w, . w 2 

But v— — , v,= , and v = . 

gP gPi gP 2 

w^w^ 

P Pi P2 

or, since their densities are as their specific gravities (150), 
w_w 1 w 2 

G ~ G l G 2 

Substituting in succession the values of ic, and w 2 , obtained 
from (6), we have 

' M , i=U) (I_i)^(l_l) = ^K W) (155) 

\a oj \o 1 aj (o 2 -o 1 )o v 

w 2 =w[ )~[ ) =)r^ } -^.w. (156) 

\o o 2 J \a 2 o l J (o x —o 2 )o v ' 

408. Hydrometer. Instruments for determining the specific 
gravity of fluids are called Hydrometers or Areometers. They 
are made of glass, brass, &c, and are of two kinds : one in 
which the weight is constant, the other in which the volume is 
constant. In its simplest form the areometer consists of a hol- 
low globe, one of whose diameters is prolonged in a flat or 
cylindrical stem of uniform size, and to the other extremity is 
attached a smaller globe loaded with mercury or shot, that it 
may float in a vertical position. 

1°. Areometer of a constant weight. 

The annexed figure may represent this form of the instru- 
ment. To graduate the stem, suppose it to sink in distilled 
water, at a given temperature, to a point s, and let the distance 
sr be divided into any number of equal parts, and continued 



266 



HYDROSTATICS. 



upward from s. When immersed in another fluid, 
suppose it to sink to t, distant from s,x divisions. Let 
V be the volume of the portions immersed in water, 
v the volume included between any two divisions of 
the stem, and o the specific gravity of the second fluid. 
Then p and p l being the densities of this fluid and 
water, the weight of the water displaced by the in- 
strument will be gP\V 9 the weight of the other fluid 
displaced will be gp(Y— vx), and, since each is equal 
to the weight of the instrument, 

gPiV—gp(y.-v%). 

V 

p, V — vx V v ' 

x 

v 

If a be previously known, and x be observed, we can de- 
termine the value of 

V a 



—=x.- . 

v I — a 

V 

Substituting in (a) this value of—, and putting for x, 1, 2, 3, 

dec, the corresponding values of o for each division of the scale 
will be known. These may be marked on the instrument, or 
the divisions may be numbered, and a table of their values 
formed to accompany the instrument. 
Cor. From (a) we have 



x=-(l — . 



Giving to a a small increment do, the corresponding increment 
of a; is 

x'—x—dx= .-^'do; 

v o 

that is, for any small increment of the specific gravity of the 

fluid the corresponding change in the depth of immersion of 

VI 

the instrument varies as —.—„, which may be considered a meas- 
v a 2 

ure of the susceptibility of the instrument. 



SPECIFIC GRAVITY. 



267 



2°. Areometer of a constant volume. 

The principal obstacle to the use of the areometer of a con- 
stant weight is the inconvenience and difficulty of calculating 
and marking against the different divisions of the stem of each 
instrument a different scale of specific gravity, and construct- 
ing the stem of that perfectly uniform thickness which is nec- 
essary to the accuracy of the observations. 

To obviate these difficulties, Fahrenheit conceived the idea 
of sinking the instrument always to the same depth by means 
of weights to be placed in a cup at the end of the stem. 

Let W be the weight of the instrument, w^ and w the weights 
respectively necessary to sink it to the same point 
a of the stem in water, and in the fluid whose 
specific gravity o is required, V the constant 
volume of the portion immersed, p x and p the 
densities of the water and the fluid. Then 
gpY=W+w and gp 1 V=W+w l . 
_ p _ W+w 

Cor. Differentiating, we get 

dm = ( W + w l ) do =gp , Yda ; 
that is, for any given small variation da in the 
specific gravity of the fluid, the variation of w is as V, or the 
susceptibility of the instrument is as the volume of the portion 
immersed. 

409. Nicholson } s Hydrometer. This is a modification of the 
preceding, to adapt it to the determination 
of the specific gravity of solids as well as 
liquids. For this purpose, a metallic basket 
is attached to the lower extremity, in which 
the body may be placed, and its weight in 
water ascertained. The basket admits of 
reversal, so that the body may be retained 
under water when specifically lighter. 

Let w be the weight in C necessary to 
sink the instrument to/. Replacing w by 
the body, let w l be the weight which must 





268 HYDROSTATICS. 

be added to sink the instrument to the same depth/. Removing 
the body to the basket beneath, let w 2 be the weight in C req- 
uisite to sink the instrument a third time to/. 

Calling W the weight of the instrument, W, that of the 
water displaced by it when immersed to f, V the volume of 
the body, p its density, that of water being p , . 

Then W+ii;=W lf (a) 

gpV+W+to^W,, (b) 

and gP Y+W+w 2 =W l -\-gp l ~V. (c) 

Subtracting (a) from (b), and (b) from (c), 
gpV=w— w xi 
gPi Y=w 2 -w 1 . 
p w—w, 

.*. G= = -. 

p, w 2 —w 1 

410. EXAMPLES. 

Ex. 1. A piece of wood weighs 12 lbs., and when annexed 
to 22 lbs. of lead, and immersed in water, the whole weighs 
8 lbs. The specific gravity of lead being 11, required that of 
the wood. 

The specific gravity of lead being 11, if v be its volume, 
llu=22 or v=2. But this is the volume of water displaced 
by the lead, and the specific gravity of water being 1, its 
weight will be 2. Therefore the loss of weight in water is 
2 lbs., and the actual weight in water 20 lbs. 

Hence, Art. 405, w=l2, w'{ = 8, and ^=20. 

_ w _ 12 _ X 

"'" °~ w+w\—w'l~ 12+20-8~2* 

Ex. 2. A diamond ring weighs 69| grains, and when weigh- 
ed in water 64| grains. The specific gravity of gold being 16 J, 
and that of diamond 3J, what is the weight of the diamond ? 

Let i?, v' be the volumes of the gold and diamond respective- 
ly, and A the weight of a unit of volume of water. Then vX is 
the weight of a volume of water equal in bulk to the gold, and 
3 2 3 vk the weight of the gold. In the same manner, \v'X is the 
weight of the diamond. Hence 



SPECIFIC GRAVITY. 269 

m\=\*vX+iv'X, (a) 

and Q4\=\*vX+iv'X-(v+v')X. (b) 

Subtracting (b) from (a), we have 
&=Xv+Xv'. 
.: from (a), 69|= y(5-At>')+}Ai>', 

139=165-26A»', 

or the weight of the diamond is three and a half grains. 

Ex. 3. A body A weighs 10 grains in water, and a body B 
weighs 14 grains in air, and A and B connected together weigh 
7 grains in water. The specific gravity of air being .0013, 
required the specific gravity of B, and the number of grains 
of water equal to it in bulk. 

Let A', A" be the number of grains in the volumes of water 
equal to the volumes of A and B respectively, and a', a" their 
specific gravities. Then, by the conditions of the question, 

(a'- 1)^=10, (a) 

(a"-.0013)A" = 14, (b) 

(o>-1)ai + (o"-1)X" = 7. (c) 

From (a), (b), and (c), we have 

14(<j"— I) 

10-1 ^ -=7 

a"-.0013 ' 

^(l-^^S^-.OOlS). 

.*. a"=.8237. 
Hence, also, from (b), 

A "=^ L 0013=^4= 17 -° 23 g rainS - 
Ex. 4. When 73 parts by weight of sulphuric acid, the spe- 
cific gravity of which is 1.8485, are mixed with 27 parts of 
water, the resulting dilute acid has a specific gravity equal to 
1.6321. Required the amount of condensation which takes 
place by the mixture. 

Let A' be the number of parts by weight in a quantity of 
water equal in volume to that of the sulphuric acid, and X" in 
a quantity of water equal in volume to that of the mixture. 

Then 1.8485^ = 73, 

or A'=39.4915, 



270 HYDROSTATICS. 

and 1.6321A"=73+27=100 

or A"=61.2707. 

Now the condensation of the mixture will be expressed by the 
ratio of the diminution of the volume of water and acid, when 
mixed, to their united volume before mixture. 

A' +27 -A" 



Hence condensation^ 



A' +27 
5.2208 



0.0785. 



66.4915 

Ex. 5. Two fluids, the volumes of which are v and v', and 

specific gravities o and a', on being mixed, contract -th part of 

lb 

the sum of their volumes by mutual penetration. Required 
the specific gravity of the mixture. 

Let a" be the specific gravity of the mixture ; then the sum 
of their weight before and after mixture will be equal. 

Hence (vo+v'o>)X=(l- \v+v')o"X, 

.... n vo+v'o' 

which gives a"= -. ; — — . 

n— 1 v-j-v' 

Ex. 6. A body whose weight in a vacuum is 73.29 grains 
loses 24.43 grains by immersion in water. Required its spe- 
cific gravity. Ans. o=3. 

Ex. 7. Required the specific gravity of a body which weighs 
65 grains in a vacuum, and 44 grains in water. 

Ans. (7=3.0952. 

Ex. 8. An areometer sinks to a certain depth in a fluid whose 
specific gravity is .8, and when loaded with 60 grains it sinks 
to the same depth in water. What is the weight of the in- 
strument? Ans. w= 240 grains. 

Ex. 9. A compound of gold and silver weighing w=10 lbs. 
has a specific gravity 0=14, that of gold being tf'=19.3, and 
that of silver being <7"=10.5. Required the weight w' and w" 
of the gold and silver in the compound. 

Ans. w' = 5.483, 
^"=4.517. 



CHAPTER III. 



COMPRESSIBLE OR AERIFORM FLUIDS. 

411. Prop. To find the tension of the atmosphere or other 
compressible fluid. 

Let a glass tube AB, open at one end and a 

closed at the other, be filled with mercury. 
Retain the mercury in the tube by the press- 
ure of the finger, let it be inverted and the 
open end immersed beneath the surface of the 
mercury in the vessel CDFE. It will now be 
found that a column of mercury, as BL, will 
occupy a portion of the tube, while the re- 
maining portion AL will be void. This col- 
umn of mercury is sustained in the tube by the 
pressure of the atmosphere on the surface of 
the mercury in the cistern CF, and as there is 
no pressure on the mercury at L, the equilib- 
rium is due to the equality of pressures of the 
atmosphere and mercury on the common base 
B (Art 401). If we now suppose the cistern 
covered, so as to separate the air without from 
that within the cistern, the pressure of the ex- 
ternal air can not be communicated to that 
within, and the mercurial column must be sustained by the ex- 
pansive force or tension of the inclosed air, and be a measure 
of it. 

If the tension of any other elastic fluid inclosed in a vessel 
be required, let the tube I from the cistern CEFD be fitted to 
an aperture in the vessel, and a communication be thus estab- 
lished between the inclosed fluid and the mercury in the cis- 
tern. The mercury in the tube will then rise or fall till an 
equilibrium takes place between the expansive force of the fluid 




272 HYDROSTATICS. 

and the weight of the mercurial column. The height at which 
the mercury stands in the tube above the surface of mercury 
in the cistern is ascertained by a graduated scale attached to 
the tube. 

412. Schol. The mean pressure of the atmosphere at or 
near the level of the sea is generally employed in mechanics 
as the unit of pressure, and other expansive forces are com- 
pared with this and expressed in atmospheres. It has been 
ascertained by the barometer that the mean pressure of the 
atmosphere, at a temperature of 50°, is equivalent to a column 
of mercury 30 inches in height ; or, the specific gravity of mer- 
cury being 13.598, to a column of water 13.598X2.5 feet=34 
feet ; or, the specific gravity of air, at a temperature of 50°, 

being — — , to a column of air of uniform density, 27,540 feet 

= 5.2 miles in height. 

The tension is also measured by the pressure of the atmos- 
phere on a unit of surface. Now 30 cubic inches of mercury 

is equal to 13.598X30 cubic inches of water =407.94Xr— t; 

1728 

cubic feet of water=0.23607Xl000 oz. = 14.75 lbs. on the 
square inch. 

The instrument above described involves the essential parts 
of a barometer, a more particular description of which belongs 
to Physics. Generally, instruments employed to determine the 
tension of elastic fluids are called manometers. 

413. Prop. To show that the tension of an aeriform fluid is 
inversely as its volume. 

Let AB be a vertical tube which communicates with the 
cylindrical vessel DEOjH!, closed at the top, and mercury be 
carefully introduced into the tube, so as to fill the horizontal 
part HjOjBC, and leave the air in DO T of the same density 
as the exterior air. The inclosed air will then have a tension 
of one atmosphere. Take DH a = |DH,, DH 3 = |DH 1 , &c, 
and draw the lines H 2 2 R 2 , H 3 3 R 3 , &c. Now if mercury 
be poured into the tube AB until it stands at H 2 2 in the 
cylinder, it will be found to stand in the tube at the height 



COMPRESSIBLE OR AERIFORM FLUIDS. 



273 



L 2 R 2 =30 inches above H 2 2 ; or, when the air 
in the cylinder is reduced to half its volume, its 
tension or expansive force is two atmospheres. 
When the mercury rises in the cylinder to the 
height H 3 3 , it will be found to stand at the 
height L 3 R 3 =60 inches above H 3 3 ; or, when 
the air is reduced to one third of its volume, it 
has a tension of three atmospheres, or, general- 
ly, the tension is inversely as the volume. The 
truth of the law has been tested experimentally 
as far as 26 atmospheres, and for all fractions 
of an atmosphere. 

If V and V be the volumes of a given mass h* 
of air, p and p' the corresponding tensions or Hz 
pressures on a unit of surface, then 



D 

m 



04- 

"i0 3 

02 



V V 

£-= Y ,ovpV=p>V>. 



Hi 



(157) 



U 



L2 



R4 
S3 



Ri 



This law is called the law of Mariotte, from the discoverer. 

414. Cor. Let p and p' be the densities of any mass of air 
corresponding to the volumes V and V. Then, since the 
density is inversely as the volume, 

p V p 



p'~V~p" 



(158) 



or the elastic force is directly as the density. 

415. Prop. To estimate the effect of heat on the volume and 
tension of atmospheric air. 

When air is inclosed in a vessel and heat is applied, its elastic 
force is increased, as may be shown by the method indicated 
in Art. 411, and by the same method the increase in the ten- 
sion for a given increase of temperature may be ascertained. 
Experiment indicates that the tension of a given volume of 
dry air increases, by being heated from the freezing to the 
boiling point, 0.367 of its original value, and therefore, if the 
tension remains the same, the volume will increase 36.7 per 
cent. Let v be the increment of volume, the tension remain- 
ing the same, for one degree of Fahrenheit's thermometer; then 

S 



274 HYDROSTATICS. 

v=0.002039, and for an increase of t degrees of temperature, 
the increase of volume will be v*= 0.002039*. 

If now V be the original volume of a given mass of air, and 
it be heated t x degrees, the tension remaining the same, the 
new volume V, will be 

and when heated * 2 degrees the corresponding volume will be 
V 2 = (1+** 2 )V . 
V, 1+itf, 1+0.002309*, 



V 2 l+vt a 1+0.002309*. 



(159) 



But the densities p l and p 2 are inversely as the volumes. 
Hence 



P, V 2 1+v*, 



p 2 v, i+vt 1 

If, also, a change take place in the tensions at the same time, 
let jo be the tension at 32°, p x the tension at 32 +* 15 andp 2 that 
at 32°+* 2 . Then, since the tension is inversely as the volume, 

V.^l+W^V, and V 3 = (1+W 2 )^V . 

. v,_ (i+^,K 

"V 2 (1 +*,)/>,' 

in which &, and b 2 represent the measures of the tensions p l 
and p a9 or the corresponding heights of the barometer. 

416. Prop. To find the density of the air at different tempera- 
tures and under different pressures. 

By accurate experiments the weight of a cubic foot of air at 
a temperature of 32°, when the barometer stands at 30 inches, 
is found to be p=0.08112 lbs. avoirdupois. Hence, for the tem- 
perature 32°+*°, 

p 0.08112 „ , ■ 

^ = T+7ri+0.002039* IbS * ( 161) 

If, also, the barometer, instead of &=30 inches, should stand 
at some other height, as b x inches, the density will be express- 
ed by 

P ft 1= 0.08112 b 1== 0.002704ft 

Pl ~l+vt'b~ 1+0.00204**30" 1+0.00204* S * * ' 



COMPRESSIBLE OR AERIFORM FLUIDS. 



275 



Whenever the elasticity of the air is expressed by the press- 
ure p— 14.75 lbs. on the square inch, instead of the barometric 
height b, the density for any other tension p , will be 

0.08112 p 1 0.0055;?, 



P P 



lbs. (163) 



ri l+vt p 1+0.00204U4.75 1+0.002042 

The density of steam is five eighths of the density of atmos- 
pheric air for the same temperature and tension. Therefore 
we have, for steam, 

0.00344p, 



1+0.002042 



lbs. 



H 



O 



417. Prop. To determine the height corresponding to a given 
density of the atmosphere, and, conversely, the density in terms 
of the height. 

Since the density of the atmosphere at the surface of the 
earth is due to the pressure of the superincumbent portions of 
it, the density must decrease as the height increases. 

Let HOBA be a vertical column of air whose 
base AB is one square foot. Conceive it divided 
into portions of equal weights w, and heights x , x , , 
x 2 , &c, beginning at AB, so small that the den- 
sity of each may be regarded as uniform. Then 
x +x x +x 2 , &c. . . . x n =2.x= h is the height of 
the nth stratum. Let p be the tension of the 
lowest stratum, and p n that of the rath, p and p n 
their densities respectively. Then the weight of 
the nth stratum is 



w =l, XnPn =^£lPL\ (158 ). 

Po 

Po'Pn 



(a) 



But the tension of the (n— l)th stratum is p n +w, and the 
height, therefore, 



w 



Po'Pn+w 



276 




HYDROSTATICS. 






In like i 


manner 


'» ^71-2 


Po' 


w 
'p n +2w' 


.Po 
Po 






X Q = %n— n 


Po 


w 

' p n +nw 


w 


since p„ 


-\-nw— 


Po' 










Now. 


, in the 


exponential 


series, 







when y is very small, we have 

e*=l+y, 
or y=yle=L(l+y)=2.30258L.(l+y), (b) 

in which I denotes the Naperian logarithm, and L the common 
logarithm of a number. 

w 
If in (b), for y, we put — , we get 

w 

p n * py 

•\ from (a), we have 



=/.(i + ^)=/.(^)=/. (i ,„ + ,)-^. 



Po 

Now, substituting jo n +w successively for p n , we obtain 

Po 

x n _ 1 =^{L(p n +2w)-L( Pn +w)}, 

Po 
Po 



x =x n ^=^{L(p n +nw)-L(p n +(n-l)w)\. 

Po 

By taking the sum of these equations, the terms in the 
brackets will all cancel, except L(p n +nw)=l.p in the last 
and l.p n in the first, and we shall have 

s , a . =A= |». (ii , o _^)=£»?.|«. (]64 ) 

r o Ho Pn 



COMPRESSIBLE OR AERIFORM FLUIDS. 277 

To find p n when h is given, we have 

Pn Po 

or £*=*'• . 

Pn 

Hence p n =p Q .e Po , (165) 

where e=2.71828, the base of the Naperian system of loga- 
rithms. 

418. Schol. Formula (164) may be adapted to the determ- 
ination of heights by the barometer. To this end let b and b n 
be the heights of the barometrical columns at the lower and 

p b 
upper stations respectively. Then, since — == ~ 9 

£=2.30258^.L^. (a) 

The value of — may be determined from the consideration 
Po . 
that p expresses the weight or pressure of a column of the at- 

mosphere on a unit of surface as one square foot, and — - must 

P e 

express the height of this column on the supposition that its 

density is uniform. Now a cubic foot of air at 32° weighs 

0.08112 lbs., and a cubic foot of water at the same temperature 

weighs 62.37917 lbs., and therefore the specific gravity of 

62.37917 
water referred to air is - * =769. Hence the height of 

a homogeneous atmosphere at a temperature of 32° is 34X769 
=26146 feet. 

.-. 2.30258^=2.30258X26146=60204 feet. 

Po 

It is here assumed, however, that the lower station is at or 
near the level of the sea, and no account is taken of the varia- 
tion of gravity at different elevations. From numerous obser- 
vations made at different elevations above the sea, and at known 
differences of height, this coefficient is found to be 60345 feet 



278 HYDROSTATICS. 

at a temperature of 32°. But the actual temperature of the air 
at both the lower and upper stations will, in general, differ from 
the standard temperature of the formula, and, since the density 
of air varies uniformly with the temperature, we may use the 
mean of the temperatures of the air at the two stations. Let 
t and t n be the indications of Fahrenheit's thermometer; then 
the mean temperature will be %(t +t n ), and the deviation t from 
the standard will be 

*=K'o+0-32. 

The expansion of dry air is 0.00204 for a change of 1°; but 
when the atmosphere contains vapor, it is found, by comparing 
the rates of expansion of vapor and of dry air, and assuming a 
certain mean humidity for the air, that the rate is expressed by 
0.00222. Incorporating this correction in (a), and using the 
coefficient determined by observation, we have 

/i=60345(l +.00222*)L.-A (166) 

It is further obvious that a change in the length of the mer- 
curial column will be produced by a change of temperature of 
the mercury. Let r and r n be the temperatures of the mer- 
cury, as shown by a thermometer attached to the cistern of the 
barometer; then, since mercury expands at the rate of 0.0001 
for each degree, 

^^(l+.OOOl)^-^), 

where b' n is the observed height of the barometer at the upper 
station. Using this value of b n , the difference of elevation be- 
tween two stations or the height of a mountain may be determ- 
ined with considerable accuracy. 

In the determination of the constant coefficient, the variation 
of gravity at different elevations is allowed for in the assump- 
tion that this coefficient is that which belongs to the mean 
height above the sea at which observations are usually made, 
and to the latitude of 45°. When the latitude differs from this, 
it will be necessary to multiply the result by 

(1 +.002837 cos. 2V>), 
V> being the latitude at which the observations are made. 



COMPRESSIBLE OR AERIFORM FLUIDS. 279 



419. EXAMPLES. 

Ex. 1. A cylindrical tube 40 inches long is half filled with 
mercury, and then inverted in a vessel of mercury. How 
high will the mercury stand in the tube, the pressure of the ex- 
ternal air being equivalent to 30 inches ? 

Let / be the length of the tube, a the length of the portion 
occupied by the air before it was inverted, h the height of mer- 
cury due to the pressure of the external air, x the height of the 
mercury after the tube is inverted, and h' the column of mer- 
cury equivalent to the tension of the air in the tube. 

Then (157), ha=h'(i-x). 

But h=h'+x or h'=h—x. 

.'. ha=(h—x) (l—x), 
whence we get x=\{l-\-h)±\ V(l— xy+4ah. 

Using the data of the question, Z=40, a=20, and 7^=30, we get 

a;=10 or 60. 
The first value is that which pertains to the specific question. 

Ex. 2. A tube 30 inches long, closed at one end and open at 
the other, was caused to descend in the sea w T ith the open end 
downward until the inclosed air occupied only one inch of the 
tube*! How far did it descend ? 

Ex. 3. A spherical air-bubble having risen from a depth of 
1000 feet in water, was one inch in diameter when it reached 
the surface. What was its diameter at the bottom ? 

Ex. 4. Required the equation of the curve described by the 
extremities of a horizontal diameter of the air-bubble of Ex. 3, 
supposing its center to move in a vertical line. 

Ex. 5. What number of degrees must a given volume of air 
be heated to double its elasticity ? 

Ex. 6. The following barometrical observations were made 
at the White Rocks, on the bank of the Connecticut River, 
some two miles below the city of Middletown : 



Barometer. 

At the base, b = 30.09 in. 


Det. Ther. 

* =83.0 


At. Ther. 

r =84.5 


On the summit, b' n =29.65 in. 


t n =85.0 


T n =83.5 



280 HYDROSTATICS. 

Hence t=±(t +t n )-32°=52° and 1 +.00222*= 1.1 1544. 

Also, r -T n =l° and 6 n =6i(l+.OOOlXl) =29.653 in. 

T 30.09 
.*. A=60345X1.11544.XL- 



29.653 
29.653 a.c. " 8.5279314 



30.09 log. 1.4784222 



0.0063536 log. 7.8030199 
1.11544 " 0.0474462 
60345. " 4.7806413 



427.67 feet " 2.6311074. 

Ex. 7. The following observations were made by Humboldt 
at the Mountain of Quindiu, New Grenada, in lat. 5°: 

Barometer. Det. Ther. At. Ther. 

At the level of the Pacific, 30.036 in. 77°.54 77°.54 

On the summit, 20.0713 65°.75 68°.00 

Required the height of the mountain. 

Ans. 11500 feet nearly. 



HYDRODYNAMICS. 



420. Prop. The velocity of a fluid in a tube of variable di- 
ameter, kept constantly full, is in different transverse sections in- 
versely as the areas of the sections. 

Since the tube is supposed constantly full, and the fluid in- 
compressible, the same quantity of fluid must pass through 
every section in a unit of time. But admitting the fluid to have 
the same velocity in every part of the same section, the quan- 
tity which flows through any section in a unit of time will be 
the product of the area of the section by the velocity. If, 
therefore, k and k' be the areas of any two sections, and v and 
v' the velocities at each respectively, 

kv=k'v', 
and v : v'=k' : k. (167) 

In reality, the velocity is diminished by the sides of the tube, 
and is therefore in any section least near the sides of the tube 
and greatest near the central portions. 

421. Prop. The velocity with which a fluid issues from a small 
orifice in the bottom of a vessel kept constantly full, is equal to 
that acquired by a body falling freely through the height of the 
fluid above the orifice. 

Let EF represent a very small orifice in A lm b 
the bottom of the vessel ABCD, filled with a 
fluid to the level of AB, GF a stratum whose 
thickness FH=h' is indefinitely small, and 
FM=A the whole height of the column ver- 
tically above the orifice. 

If now the stratum GF fall by its own 



D 



G- 



weight through HF=A', the velocity will be E F 



282 HYDRODYNAMICS. 



v— V2gh'. 
But if the stratum be urged by its own weight and the weight 
of the column LH, calling the force in this case g', the velocity 
v 1 will be 



V '== V2g l h'. 
But the forces g and g' are as the weights of the columns 
GF and LF, or as their heights h' and h. Hence 
s' h , gh 

Substituting this value of g' in that of v', we have 

v'=V2gh, (168) 

which is the velocity of a body falling freely through the 
height h. 

Cor. If the orifice be made in the side of the vessel, and a 
tube be inserted so as to direct the current obliquely, horizon- 
tally, or vertically upward, the velocity of efflux will be the 
same, since the pressure of fluids is the same in every direc- 
tion. 

In the first case, its path will be a parabola whose equation 
is (64). 

In the second case, the angle of elevation a=0, which re- 
duces the equation to 

x 2 =4hy, 
the equation of a parabola whose axis is vertical and origin of 
co-ordinates at the vertex or orifice. 

In the last case, if all obstructions are removed, the fluid will 
rise to the height of the surface of the fluid in the vessel. 

422. Prop. To determine the horizontal distance to which a 
fluid will spout from an orifice in the vertical side of a vessel. 

Let the vessel ABCD be filled to the level AB. If the fluid 
issue horizontally from the orifice O, the equation of its path is 

x 2 =4hy, 
in which h— OB, the height of the fluid above the orifice, is the 
impetus or height due to the velocity. 

To determine the range, let y=OC—a. Hence 



DISCHARGE FROM SMALL ORIFICES. 



283 




x=CY=2Vhy=2Vha-. 
or the horizontal distance is 
equal to twice the ordinate at 
the orifice, in a semicircle whose 
diameter is the vertical distance 
from the surface of the fluid to 
the horizontal plane. 

Cor. When the orifice is at 
the middle of BC, the range is a 
maximum, and at equal dis- 
tances above and below the mid- 
dle the range will be the same. 

423. Prop. To find the quantity of fluid discharged in a 
given time from a small orifice in the bottom of a vessel when 
the fluid is maintained at the same constant height. 

Let h be the constant height of the fluid in the vessel, t the 
given time, k the area of the orifice, and Q, the quantity. 

Then Q=kvt=ktV2gh. (169) 

If the orifice be circular, and r its radius, then k=7rr*, and 
Q=n V~2g.tr* Vh, 
=25.105.*rVX 
The time being in seconds, and r and h in feet, Q, will be in 
cubic feet. If the weight W be required, 

W = 62.5Q. 

Cor. 1. The time required for the efflux of a given quantity is 

-nrW2gh 
Cor, 2. Since t and g are constant in (169), 
QackVl. 
Hence the quantity discharged in the same time from orifices 
differing in size and distance from the surface, varies as the 
size of the orifice and square root of its depth. 

If the orifices are the same, QocVA. In order, therefore, 
that the discharge from one orifice may be twice that from an- 
other, its depth must be four times as great. 



284 HYDRODYNAMICS. 

424. Prop. To determine the time in which a cylindrical ves- 
sel filled with water will empty itself by a small orifice in its 
base. 

The velocity of efflux at any instant is v= V2gh, h being the 
height of the fluid in the vessel. But since the velocities v and 
v', at the orifice and at any transverse section of the vessel, are 
inversely as their areas, k and k' (167), 

fc ^ _ 

v '—Ti v= tj V2gh ac Vh. 

The surface of the fluid, therefore, in the vessel descends with 
a velocity proportional to the square root of the space over 
which it must pass, and the circumstances of its motion are 
precisely the same as those of a body projected vertically up- 
ward. But the velocity of a body projected upward is such 
that if it continued uniformly it would move over twice the 
space through which it must move in the same time to acquire 
or expend that velocity. Hence the time of descent of the 
surface of the fluid to the base is twice that required for the 
descent of the same superficial stratum when the vessel is kept 
full. In the latter case, if Q, be the whole quantity of water 
in the vessel (169), 

kV2gh 
Therefore, in the former, 

f=-S=. (170) 

kV2gh 

If r' be the radius of a tranverse section of the vessel, r that 
of the orifice, 

Q=nr' 2 h and k=nr*. 

.-. t= =Vh. 

rW2g ■ 

425. Schol. The preceding deductions are founded on the 
hypothesis that the fluid particles descend in straight lines to 
the orifice, and all issue with a velocity due to the height of 
the fluid surface. Experiment shows that this is true only of 



DISCHARGE FROM SMALL ORIFICES. 285 

those vertically above the center of the orifice, that those sit- 
uated about the central line of particles, take a curvilinear course 
as they approach the orifice, being turned inward toward this 
line or spirally around it, and this deviation from a vertical 
rectilinear path is the greater the more remote they are from 
the central line. This deviation will necessarily occasion a 
diminution of vertical velocity, and therefore a diminution of to- 
tal discharge. The oblique direction of the exterior particles 
gives to the vein of issuing fluid, when the orifice is circular, 
the form nearly of a conic frustum, whose larger base is the 
area of the orifice. This diminution of a section of the issuing 
fluid is called the contraction of the vein, and the vein itself, 
from the orifice to the smallest section, the vena contracta, or 
contracted vein. 

The results of most experiments agree in making the length 
of the contracted vein, when the orifice is circular and hori- 
zontal, equal to the radius of the orifice. There is a greater 
discrepancy in the results of experiments for determining the 
ratio of the diameters of the two ends of the contracted vein. 
When water flows through orifices in thin plates it is found to 
be about 0.8. The ratio of the areas of the two ends, called 
the coefficient of contraction, is therefore 0.64. 

The actual discharge is found by experiment to differ slight- 
ly from the theoretical for other causes. The ratio of the 
former to the latter is found to be about 0.97, and is called the 
coefficient of velocity. The product of the coefficients of ve- 
locity and contraction, called the coefficient of efflux, is 0.62. 

Hence, for the actual velocity of discharge through orifices 
in thin plates, we have 

v 1 =0.62^=0.62 V~2gh, 
and for the quantity in a unit of time (169), 

Q^=kv 1 =0.62kV2gh. 
When the orifice or pipe through which the discharge is made 
has the length and form of the vena contracta, the velocity 
will be v , =0.97^=0.97 V~2gh, 

and the quantity = ^, = Q.91kV2gh, 
k being the smallest section of the contracted vein. 



286 



HYDRODYNAMICS. 



By the use of cylindrical or conical adjutages, the quantity 
of the discharge is increased. More seems to be gained by 
the adhesion of the fluid particles to the sides of the tube, in 
preventing the contraction of the vein, than is lost by the fric- 
tion. The discharge is greater when the adjutage is conical 
and the larger end is the discharging orifice. 

426. Prop. To determine the quantity of water which will 
flow from a rectangular aperture, 

1°. When one side of the aperture coincides with the sur- 
face of the water. 

Let h be its height and b its breadth, and 
conceive the aperture to be divided horizon- 
tally into a very large number n of equal di- 
visions so narrow that the velocity of the 
fluid in every part of each may be regarded 
as the same. The larger n is, the more near- 
ly will the hypothesis be satisfied. 

The depth of these successive divisions be- 



B 



low the surface will be 



h 2h 3h 
-, — , — , &c. 
n n n 



The velocities in each will be 

/~~h /~~2h / 3h - 

and since the areas of these divisions are each &-, the quanti- 

n 

ties discharged by each in a unit of time will be (169) 
bh /~~h bh I 2h bh /~~Sh . 

and their sum, or the whole quantity Q, discharged, will be 



But 



nVn 



l*+2 2 +3 2 +, &c. 



l+i 



= |7i a . 



DISCHARGE FROM RECTILINEAR APERTURES. 287 



Q=!LbhV2gh=%bV2gh 3 



bh 



Or, if u=77 be the mean velocity, 

v==%V2gh. 

2°. When the upper side of the rectangular aperture is not 
coincident with the surface of the fluid. 

Let its depth EL=/ij, and the distance EG=^, as before. 
Then the quantity which issues from the aperture LH will be 
equal to the quantity which would issue from EH, diminished 
by the quantity which would issue from EK, or 



Q=|6 V2g-A'-|6 V2ghl=ibV2g(h 2 -hl) ; 
and if Q=b(h— h x )v, 

3 3 

427. Prop. To determine the quantity of water which will 
flow from a triangular aperture. 

1°. When the vertex of the triangle is in the surface. 

Let the base GH=6, and the height HF=h, E _ 
and let the triangle be divided into n parts, as 
before, of equal but very small heights. The 



altitudes of these elements are each -, and the 

n 



H 



lengths, beginning at the vertex of the trian 

b 2b 3b D ^ . 
gle, are -, — , — , &c. 1 heir areas, regard 
° n n n ° 

ing them as parallelograms, which we may do, since n is in 

definitely large, are 

bh 2bh Sbh 
-.-, — .-, — .-, &c. 
n n n n n n 

Hence the discharges through each are 

bh f~h 2bh /~~2h Sbh /~~3h ' 
71 V b n ?i a V & n tf V ° n 

and the whole discharge will be 



288 



HYDRODYNAMICS. 



' bhV2gh 333 
Q=— ~(l 2 +2 2 +3 2 +, &c. 



3 



MV2gh=%bV2gh\ 



nWn 
__ bhV2^h /J +1 \_ 

Also, «=— =i^/2^. 

2°. When the base of the triangle is in the surface. 

In this case the quantity which will flow through EFG will 
equal the discharge through EFGH minus that through FGH, 
or Q = %bh V~2gh-%bh V~2gh= T %bh V~2gh, 

and v= T *jV2gh. 

Cor. If the aperture be a trapezoid, as 
EFGH, whose upper base EF=6j lies in 
the surface, lower base HG=b 2 , and alti- 
tude KH=A, we may find the discharge by 
regarding the aperture as made up of the 
rectangle KHGM, and the two triangles 
EKHandMGF. Hence 

Q=|& 2 7iV r ^+ T \(& 1 -6JV2P= T \(26 1 +3& 2 )AV2p. 
Also, through the triangle HOG, whose base HG=& 2 , and 
depth of vertex LO=A,, the quantity of the discharge will be 

Q=A&A V2^h~ l -M^ 1 +sb 2 )hV2^h. 

In a similar manner, rectilinear orifices of other forms may 
be divided into triangles, trapeziums, &c, and the discharge 
determined. 

428. Prop. To determine the velocity with which an elastic 
fluid will issue from a small orifice into an unlimited void when 
urged by its own weight. 

By reference to Art. 421, it will be seen that the result there 
obtained is independent of the density of the fluid, and will 
therefore be true of all fluids, whatever be their density. For 
the forces g and g' are as the weights of the stratum at the 
orifice, and of the column vertically above it ; that is, 
g : g'=w : w'=gh : gh'=h : h'. 




MOTION OF FLUIDS IN LONG PIPES. 289 



Consequently, v= V2gk' 

expresses the velocity with which every fluid of uniform dens- 
ity will issue from a small orifice. 

Cor. The velocity with which air will rush into a vacuum 
is that which a heavy body would acquire in falling from the 
height of a homogeneous atmosphere. The height of a homo- 
geneous atmosphere {Art. 418), when the temperature is 32° 
and the barometer stands at 30 in., is A' = 26146 feet, which 
gives v=\2$l feet. But for a temperature of 60° the height 
h, will be (159) 

h^h'il+vt), 

in which t=60° — 32°=28°, and the value of v for a mean state 
of humidity of the air will be 0.00222 instead of 0.002039 
(Art. 412). 

.-. v= V2gh 1 = V2gh'(l+0.00222X28) 



= V2X32iX26146X(l+0.00222X28) = 1336 feet. 

MOTION OF FLUIDS IN LONG PIPES. 

429. The subject of the conveyance of water in pipes is one 
of considerable practical importance. But the motion of fluids 
in long pipes is so much affected by adhesion and friction in 
the interior, by the resistance occasioned by curvature, and by 
the disengagement of air, which remaining stationary when 
the pipes are laid along a level surface, or collecting in the 
higher portions of them, where they are curved, opposes the 
flow of the fluid, that theoretical results are of little practical 
value ; besides, investigations conducted on hypotheses in- 
volving all the causes which affect the motion of the fluid are 
too difficult for an elementary work. 

The experiments of Bossut, in 1779, are those which are 
usually relied on for information on this subject. Water was al- 
lowed to flow through pipes whose diameters were 1| in. and 
2 in., and lengths from 30 to 180 feet. They were chiefly of 
tin, and inserted in the side of a reservoir filled to a constant 
height, either one foot or two feet above the center of the pipe. 
The following principles were established : 

1°. The discharges in given times with pipes of the same 

T 



290 



HYDRODYNAMICS. 



length, and with the same head of water, are proportional to 
the squares of the diameters. 

2°. For pipes of different lengths and of the same diameter, 
the discharge was inversely proportional to the square roots 
of the lengths. 

To determine the supply which may be expected from a 
pipe of given dimensions, it was found that a pipe 30 feet long 
and 1| inch in diameter would discharge at its extremity about 
one half of that which would issue from a simple orifice, or 
short pipe of the same diameter. 

Couplet, in 1730, found that a pipe of stone or iron, 600 fath- 
oms in length and 12 inches in diameter, with a head of 12 feet 
of water, discharged T Vth, and a pipe of the same diameter and 
2340 fathoms in length, with a head of 20 feet, discharged T \th 
of that which would have been obtained from a simple orifice. 

To determine the quantity which flows through a section of 
a natural stream, it is usual to measure the breadth and the 
depth at different points of a transverse section, and find the 
area of the section. The velocity at various points of the sec- 
tion is measured by the hydraulic quadrant, or rheometer of 
some kind. The mean velocity, multiplied by the area, gives 
the quantity which flows through the section in a second. 



430. Prop. To determine generally the velocity with which a 
fluid will issue from an orifice of any size in the bottom of a 
cylindrical vessel. 

A We shall adopt the usual hypothesis of a 

division of the fluid into thin laminae, and that 
in their descent their parallelism is preserved. 
Let the distance of the lamina pp'qq' from 
the surface A be x, and K the area of its 
surface. Then, if dx be its thickness and p 
its density, pKdx will be its mass, and gpKdx 
its weight, or the force with which it will 
descend if free. Now the resistance with 
which it meets in its descent is the difference 
of pressures on its lower and upper surfaces. 
help represent the pressure on a unit of the 






I) B E 



GENERAL METHODS. 291 

upper surface by the water above it ; then p+dp will be the up- 
ward pressure on a unit of the lower surface, or the resistance 
of the water below it, and the difference — dp, multiplied by 
K, will be the resistance experienced by the whole lamina. 
Hence the moving force will be 

gpKdz— Kdp, 
and the acceleration, [VI.] and [IX.], 

gpKdx—Kdp _gpdx—dp _d 2 x 
^~ pKdx ~ pdx ~~W {a) 

Let v be the velocity of discharge at B, k the area of a sec- 
tion of the issuing fluid, and dt the element of time in which 
the surface K descends a distance equal to dx. Then (167) 
we have 

T . _ _ _ dx kv 

&dx=kvdt or "^t=t7-- (o) 

By differentiation, regarding dt as constant, we get 
d z x__kdv 

W~~W3$ m 

kdv _gpdx—ap 

Kdt pdx 

pkdv dx 
g P dx-dp=^. Tt . 

k 2 
Hence (b) gpdx—dp—p^- 2 .vdv. 

k 2 v 2 
Integrating, gP^-p=P^-^+C 

Let now x=h= AB and A— K ; then 

v 2 
gph-p=p-+C ; 

and subtracting the preceding from this, we have 



V=s /M^) 



or «=V » • < 171 ) 

1- 



Iv 



292 HYDRODYNAMICS. 



If k be very small, v= V2g(h—x) nearly; and if the vessel be 
kept constantly full, x=0 and v= V2gh (168). 

If &=K, v= oc, or the velocity must be infinite. From which 
we infer that a section of the issuing fluid can never be equal 
to a section of the vessel. If a cylindrical tube be vertical and 
filled with a fluid, the portion of the fluid at the lower extremi- 
ty, being urged by the pressure of all above it, will neces- 
sarily have a greater velocity than those portions which are 
higher, and therefore (167) a section of the issuing fluid is nec- 
essarily less than a section of the tube. 

431. Prop. To determine the time in which a cylindrical ves- 
sel will exhaust itself by a small orifice in the base. 

Since k is very small, we have v= V2g(h—x) for the veloc- 
ity at the end of the time t when the surface of the fluid has 
descended through the space x. Let this velocity be sup- 
posed constant during the indefinitely small time dt. Then 
the quantity discharged in this time will be 

dQ=kdt V2g(h-x) = Kdx, 

dx being the descent of the surface in the time dt, 

K _i 

Hence dt= — =.{h—x) 2 dx, 

kV2g 

2K i 

Integrating, t= -=.(h—x) 2 +C. 

kV2g 

2K - 
If x =0, *=0, and C= — —Vh. 
kV2g 

2K 



(Vh-Vh-x), (172) 

kV2g 

and when x—h, or the whole is discharged, 

i=? K^ = jK L= ^ =(170) (m) 

kV 2g kV2gh kV2gh 

The time is therefore twice as great as that which is required 
to discharge the same quantity when the vessel is kept con- 
stantly full. 



EXAMPLES. 293 

432. Prop. To determine the quantity which will issue from 
an aperture of any size and form in the side of a prismatic 
vessel. 

Let the Fig., Art. 430, represent one face of the vessel, and 
DCE the aperture. 

The element of the area of the orifice will now be ydx. 

.'. dQl=tydx V2g(h 1 +x), 
in which h 1 =AC, x=Cm, and y—mn. 

This, integrated between the limits x=0, and x=h—h X =CB, 
will give the quantity discharged in the time t. 

If the aperture be rectangular, y=a constants. Then 

dQ,=bt V2g(h x +x)dx, 

and Q=%btV2g(h 1 +xf+C, 

and, between the limits above named, 

Q=f^V2^ f -4). (174) 

If the orifice extend to the top of the vessel, then h 1 =0 i and 

Q= \bt V2g.h*= Ibht V~2gh, 

which is the velocity due to the depth of the center of pressure 
of the aperture below the surface of the fluid. 

In the foregoing the vessel is supposed to be kept constant- 
ly full, or the surface of the fluid to be very large compared 
with the aperture. 

433. EXAMPLES. 

Ex. 1. A vessel, formed by the revolution of a semi-cubical 
parabola about its axis, which is vertical, is filled with a fluid 
till the radius of its surface is equal to its distance from the 
vertex ; to find the time in which the fluid will be discharged 
through a small hole at the vertex. 

The equation of the semi-cubical parabola is ay*=x\ 

Hence (167) kvdt=-Kdx, 

TTX^dx 

or k V 2gxdt= — ny^dx = . 



294 HYDRODYNAMICS. 

7T I 

.\ dt= z=z.x 2 dx, 

kaV2g 

Taking the integral between the limits x=0 and x — a, we 
have 

7i i irV2a b 

t= =.%a 2 = =. 

kaV2g Ik Vg 

Ex. 2. To find the time in which a paraboloid of revolution 
whose altitude is h and parameter p, full of fluid, will empty 
itself through a small orifice at its vertex, its axis being vertical, 

3 

2rrph 2 

Ans. t= =. 

SkV2g 

Ex. 3. A conical vessel, the radius of whose base is r and 
altitude h, is filled with a fluid. Required the time in which 
the surface of the fluid will descend through half its altitude, 
the orifice being at the vertex and the axis vertical. 

7rr 2 F(2 l -l) 
Ans. t= v i . 

20kg 2 

Ex. 4. Find the times in which a fluid contained in a vessel, 
formed by the revolution of the curve whose equation is y i =a z x 
about the axis of #, will descend through equal distances h x , 
supposing a small orifice at the vertex, and the axis vertical. 

3 

_ , rra 2 h 

Ans. Jbach t= — =. 
kV2g 

Ex. 5. The horizontal section of a cylindrical vessel is 100 
square inches, its altitude is 36 inches, and it has an orifice 
whose section is one tenth of a square inch. In what time, if 
filled with a fluid, will it empty itself, allowing for the contrac- 
tion of the vein ? 

Ans. *=ir36*.5. 



HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 



MARIOTTE'S FLASK. 







D 




M 
1 


[ I 


h 




l 


A= 




H 


B = 




E 
F 


C = 





433. This piece of apparatus was devised to illustrate several 
successive positions of equilibrium of air and 
water, and variations in the velocity of dis- 
charge. In the annexed section of this flask, 
A, B, C are apertures so small that when open 
and water is issuing the air can not enter, and 
the reverse. The apertures at first are all 
closed, the vessel is filled with water, and the 
tube DF is inserted at M and filled with water 
to the height D. 

1°. Let the aperture B be opened. 

If h be the height of a column of water equiva- 
lent to one atmosphere, and x represent, in every 
case, the difference of level between the surface 
of the water in the tube and the open orifice, then the pressure 
of the air inward at B on a unit of surface will be h, while the 
pressure outward on a unit of surface will be that due to the 
pressure of the air on the surface at D, together with the press- 
ure of the column of water DE=x, or the outward pressure at 
B is h+x. Therefore the pressure outward is greater than 
the pressure inward by the difference h+x—h=x, and the 
water will issue with a decreasing velocity due to x, or 
v= V2gx. This discharge will continue till x vanishes and 
the water in the tube descends to E, on a level with B, when 
the equilibrium is restored. 

2°. Let B be now closed and A be opened. 

Now the pressure downward on the surface of the water in 
the tube at E is h. The pressure upward on the same surface 
is that due to the pressure of the air at A, and to the weight 



296 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 

of a column of water equal in height to AB— HE=x, or the 
upward pressure on E is h+x. The resultant upward press- 
ure on E is therefore h+x—h=x, and the surface of the water 
in the tube will rise till x vanishes, or till it arrives at H, when 
the equilibrium will again be restored. 

In the mean time, however, the air has entered at A and 
risen to the upper part of the vessel, where it occupies the space 
above ho. Since an equilibrium now exists, the expansive 
force of the air above ho, together with a column of water 
equal in height to LH, equilibrates the pressure h of the air on 
the water in the tube at H. If h' be the height of a column 
of water equivalent to the tension of the air above ho, and 
z/=LH represent the difference of level of the surfaces of the 
fluid in the tube and flask, we have 

h'+y=h or h'—h—y. 

3°. Let A and B be closed and C opened. 

The pressure at C inward is h, outward is h+x, represent- 
ing by x, as before, the vertical distance of the surface of the 
water in the tube from the open orifice C. Therefore the re- 
sultant of the pressures at C is h+x—h=x, and the water will 
flow from C with a velocity due to the height x. As the sur- 
face of the fluid in the tube descends, x diminishes, and the 
velocity of discharge also, till the surface in the tube arrives at 
F, at which time x becomes equal to the height of F above C. 
In this position of the fluid, the equation h'=h—y still obtains, 
y being now the distance LF. But an equilibrium not being 
established, air will enter the vessel at F and rise to ho. This 
operation will continue till y=o and h' = h. During this en- 
trance of the air, and the descent of ho to F, the value of x has 
remained unchanged, and therefore the velocity of discharge 
has remained constant. After this the value of a; will diminish, 
and the velocity of discharge also, till the water in the flask de- 
scends to the level of C. 

BRAMAH'S HYDROSTATIC PRESS, 
434. In the annexed section of this instrument L and H are 
vertical cylindrical cavities, in a solid mass of metal or other 



BRAMAHS HYDROSTATIC PRESS. 



297 




strong material. The 
diameter of H is con- 
siderably less than that 
of L, and they commu- 
nicate through an aper- 
ture N, in which is a 
valve opening into L. 
A is a strong piston or 
solid cylinder of iron 
fitting closely to the 
surface of the hollow 
cylinder, and movable 
in it. IH is a piston 
similarly applied to the 
other cavity H, and 
movable by means of a lever KP, whose fulcrum is at P. At 
O is a valve opening upward, and below it the cylinder H is 
continued downward to a reservoir of water. The lever KI 
being raised, the water ascends, as in the common pump, from 
the reservoir G into the cavity H. The lever being then press- 
ed down, the valve O closes, and the water is forced through 
the valve N into L, and, acting on the piston A, communicates 
a pressure to any substance, as books, included between its 
upper surface and a strong cross bar BC firmly connected 
with the solid cylinder DE. When the water from H is forced 
into L, its reaction closes N, and the same operation is re- 
peated. 

The pressure on the base of A (134) is to the force impress- 
ed by H as the area of the base of A is to that of H. 

Let R and r be the radii of the pistons A and H, L and / 
the longer and shorter arms of the lever KP, p the power ap- 
plied at K, and P the resulting pressure on the base of A. 
Then the force p ! , applied to the piston H, will be 



But (134), 



p'= 


ph 


p 
p>~- 


ttR 2 



298 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 
. p , R2 L R2 

r r a l I r 
This press seems to present the simplest and most effective 
of all contrivances for increasing human power. 

T R 

If>=100 lbs., T =10, and -=10, then P= 100,000 lbs. 



HYDROSTATIC BELLOWS. 

435. This instrument presents an illustration of what is term- 
ed the Hydrostatic Paradox: that fluids, 

^j however unequal in quantity, may be made 

to equilibrate. It consists of two circular 
boards, A and B, firmly connected by a 
cylindrical coating of leather or other flexi- 
ble material. CM is a tube communicating 
with the lower portion of the cylinder. 
Water being poured into the tube CM, the 



K 



M 



~~~\ boards A and B will separate, B will rise, 
\ and a weight W, very large compared with 

H that of the water in CM, may be supported 

\ on B. The fluid in CM which counter- 
( poises W is that above P, the level of B. 
Let k be the area of a horizontal section 
of the tube, K that of a section of the cyl- 
inder, or that of B, and p the pressure of the fluid NP above 
Pon k. 



Then (134) 


p k 






If <t be the specific gravity of the weight W, 
and NP=A, then 


V its volume 




gp l hk k 

. pv V 







If V be the whole volume of the fluid in the instrument, and 
H the height of the cylinder AB, then 



DIVING BELL. 299 

V'=H(K+A:)+M=H(K+A;)+(7V.g. 

Y'K-oYk 
"' K(K+A) * 

Let now the quantity of fluid be increased by v, and the cor- 
responding increment AH in the height of W will be 
K(V'+v)-oYk KV'-oYk v 



AH: 



K(K+k) K(K+k) K+k' 



DIVING BELL. 

436. The diving bell is commonly a hollow cylinder or par- 
allelopiped, one end of which is closed. It is immersed with 
the open end downward, weights being added to sink and keep 
it steady in its descent. As the vessel descends the fluid con- 
tinually exercises a greater pressure on the contained air, con- 
denses it, and occupies a greater portion of the vessel. If the 
form be as above supposed, to find the height x of that portion 
of the bell free from water, when the top is sunk to the depth 
H, let h be the height of a homogeneous atmosphere, a the 
height of the bell, p' the density of the external air, p that of 
the air in the bell, and p 1 that of water. Then the pressure p, 
on a unit of surface of air in the bell, will be 

P=g{hp' + (R+x) Pl }. 
But the unit of pressure arising from the elasticity of the intern- 
al air is gph. 

.-. g\hp l + (H+x)p 1 \=gph. 

The quantity of air in the bell being constant, and a horizon- 
tal section of it the same throughout (155) and (156), 

px=p'a, or p=p'.~. 

.\ hp f -\-E.p 1 +xp l =p f a- f 

p' P' 

or —hx+H.x-\-x 2 =—ah f 

Pi Pi 

and o being the specific gravity of air, 

ohx + Hx + x* = a ah. 



x=-±(R+ch)± Vi(U+chy+<Jah. 



300 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 




Pi 



Mi 



rw~i 



SEA GAGE. 

437. The vessel AB is perforated with holes, and has with- 
in it, fixed in a vertical position, a glass tube, hav- 
ing one end closed and the other immersed in a 
cup of mercury. A is a hollow sphere whose 
buoyancy is sufficient to raise the instrument 
when the weight W suspended at the bottom is 
detached. The instrument is allowed to sink in 
the water whose depth is to be determined, and 
by a mechanical contrivance the weight is de- 
tached when it strikes the bottom, so that the gage 
will ascend by the buoyancy of the ball. The 
height MP to which the mercury has risen in the 
tube is marked on the interior of the tube, by the 
adhesion of oil or other viscid substance placed on 
its surface. 

Let h be the height of the barometer at the surface, I the 
length of the tube above the surface of the mercury in the cup, 
x the depth of the water, and M.V=h^. If h 2 be the height of 
a column of mercury which would be sustained by the elastici- 
ty of the air in NP, then (155) 

hi 
h 2 (l—h 1 )=hl or h 2 =j — j—. 

Now the elasticity of the air in NP+ the weight of the col- 
umn MP= the pressure of the atmosphere on the surface of the 
fluid + the weight of a column of water extending to the bottom; 
or, p being the density of mercury, and p, that of the water, 
ph 2 +ph 1 =ph +p j x. 

... x =-?-(h 2 +h l -h)=oh 2 +o(h 1 -h) 
Pi 

hi 
=a- — ; — \-o(h x ~h) 



l-h 
ah 



1 



h, 



\-o(h x —h), 



in which a is the ratio of the specific gravity of mercury to 
that of the water of the ocean. 



SIPHON. 



301 





1 
B 


V 


■Br 






V' 



SIPHON. 

439. The siphon is an instrument for transferring fluids 
from one vessel V to another V in which 
the surface of the fluid is lower. It consists 
of a bent tube with one branch longer than 
the other. Suppose this tube to be filled with 
the fluid from the vessel V, and to have its 
extremities immersed in the fluids in the two 
vessels. 

Let CB=a be the difference of level of the 
fluids in the vessels, p 1 the density of the fluid, 
p the density of the surrounding air, and k 
the area of a section of the tube. The press- 
ure on the surface of the fluid in V without 
the tube by the atmosphere will be greater 
than that on the surface of the fluid in V by 
gpak. This excess of pressure acting at O upward will urge 
the fluid from V to V. At the same time, the pressure exert- 
ed by the liquid in the siphon at the level of the fluid in V 
will be greater than that at the level of the fluid in V by gp Y ak. 
This excess of pressure will urge the fluid from V to V. 
Hence the resultant pressure is equal to gak(p 1 — p), and in the 
direction DAO. The fluid will therefore flow from V to V', 
and, since p is small, it will move with a force nearly equiva- 
lent to a column of the fluid whose height is a and base k. It 
is not necessary that the longer branch should be immersed in 
the fluid of V. 

If h be the height of a column of the fluid in equilibrium with 
the atmosphere, and AB=:c, then the upward pressure on k in 
the shorter branch, at the level of the fluid, is that due to h, 
while the downward pressure is that due to x. Therefore the 
fluid is urged in the direction DAO, at A, by a force represent- 
ed by h—x. As long as x<Ji the fluid will rise to A and pass 
into the longer branch. 



302 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 




THE COMMON PUMP. 

440. AB and BC are cylinders connected together as in the 
figure, the former called the body or 
barrel of the pump, and the latter the 
suction pipe. At B is a valve opening 
upward. M is a piston accurately fit- 
ting the barrel, and movable by means 
of a rod EV and a lever EF. In the 
piston is a valve V, which opens upward. 
The suction pipe BC is immersed in the 
fluid to be raised. 

To understand the action of the pump, 
conceive the piston M to be depressed 
to B. The air in MB being thus con- 
densed, will open the valve V and escape. 
If now the piston be raised to A, the 
pressure of the external air will keep 
the valve V closed, the air in BC, by its 
elasticity, will open the valve V and dif- 
fuse itself uniformly through CA. The 
pressure on the surface of the water in 
the tube at C being thus diminished, the 
pressure of the air on the water without 
the tube will cause it to rise to some 
point P in the tube, till the weight of 
PC and the elasticity of the air in AC together shall produce a 
pressure on a section of the water in the tube at C equal to the 
upward pressure, occasioned by the external air on the water 
of the reservoir ; that is, until the tension of the air in AP, to- 
gether with the weight of PC, equal one atmosphere. The 
valve V being now equally pressed on both sides, will close by 
its own weight. Let now the piston descend again to B. The 
air in AB being condensed, will again open V and escape. By 
a second ascent of the piston to A, the air in BP will expand 
through PA, and the volume of water PC will increase in 
length till the pressure on the section at C shall again be one 
atmosphere. By repeated strokes of the piston the water will 



THE COMMON PUMP. 303 

rise to B and pass through V. When this is effected, a por- 
tion will pass through V at each descent of the piston, and be 
lifted by its ascent to the reservoir above A, and be discharged 
through the spout G. 

To determine the distance BP, and the elasticity of the air in 
BP after n strokes, let h n be the height of a column of water 
which will measure the elasticity of the air in BP, and BP=# n ; 
also, let h n+1 and x n+1 represent the same quantities after the 
(n+l)ih stroke. Let h be the height of a column of water 
equivalent to one atmosphere, AB=a the length of a stroke of 
the piston, BC=b the distance of the lower valve from the sur- 
face of the water, K be a section of the barrel AB, and k a sec- 
tion of the suction pipe BC. Then, since the tension of the air 
in BP, together with the weight of the column of water PC, is 
equal to one atmosphere, 

h n +(b— x n )=h and h n+l +(b— x n+1 )=h. 
.'. h n —x n =h—b and h n+1 —x n+l =h—b. 

Now P being the position of the surface of the water after 
the 71th stroke, and Q after the (?i-H)th, h n and h )l+1 measure 
the densities of the air in BP and BQ ; and since, by the (n+ l)th 
stroke, the air in BP is expanded through AQ (155), 

h n+1 (Ka + kx n+l ) = kh H x n , 
and substituting the value of x n+1 =h n+l +b— h, we have 
h nn (Ka+k(h n+1 +b-h))=kh n x n . 



Hence K +l = -\ j -^+&-/*}=4\/ j -^+b-h}*+4h n x n , 

and x ll+1 =l j -y+A-6}±iY/| ~y+b-h} 2 +4:Kx n . 

The length of the pipe BC is necessarily less than h, other- 
wise the weight of PC, before it becomes equal to BC, would be 
equal to one atmosphere, and, could a vacuum be produced in 
MP, no further ascent of the fluid would follow. 

To find the pressure on the piston when the water is raised 
to the point N above the piston, let it be represented by a col- 
umn of water whose height is x. The pressure upward against 
the base of M is that of a column of water equal in height to 



304 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 

h— MC. The pressure on the upper surface is /j+NM. There- 
fore the resultant pressure is the difference of these, or 
#=A+NM-/*+MC=NM+MC=NC. 
441. Sometimes, by not preserving a proper relation between 
the distances AB and AC, the water will cease to 
rise, even though BC<A. To show this, let the 
barrel and suction pipe be equal, or K=k, and the 
lower valve be at some point below B, as at C. 
j^y When the water has risen to X and the piston is 
raised to A, the tension of the air in AX, together 
with the weight of the column XC, is equal to one 
atmosphere. If now, when the piston descends to 
B, the tension of the air in BX does not become 
greater than that of the external air, the valve in 
the piston will not be opened, no air will escape 
from BX, and the water will rise no further. To 
find the point X when the air in AX, reduced to 
the volume of BX, becomes equal in tension to 
one atmosphere, let AX=x, AB=a, AC=L Then 
BX=#— a. Now when the piston is raised to A, 
the air in BX being expanded through AX, the 
h(x— a) 



tension becomes (155) 



x 
-(x—a)+l—x=h, 



and x=\l±\^ir— A.alu 

If P<4ah, the value of x is imaginary, and no point of stop- 
page exists. When l 2 =4ah, x=±l, and there is one point at 
half the distance from A to C ; and when P>4ah, there will be 
two points. 

Ifa=10, 1=36, and£=32, 

#=20 and 16. 



AIR PUMP. 



305 



v 
B 



v 



v'&k 



OLE 



R 



3JF 



AIR PUMP. 

442. B and B' are cylinders or barrels, commonly of the 
same size, in which pis- 
tons P and P', with valves 
V and V opening upward, 
are movable by means of 
rack-work, the one ascend- 
ing as the other descends. 
At the lower extremity of 
the barrels there are aper- 
tures with valves opening 
upward, and communica- 
ting by the pipe ac with the 
receiver R, from which the air is to be exhausted. 

The ascent of the piston P tends to produce a vacuum in B, 
the valve V being closed by the pressure of the external air. 
The air in R, by its elastic force, opens the valve v and fills 
the barrel B. When P descends, the elasticity of the air in B 
closes v and opens V, through which the contents of B escape 
into the external air. The action of both pistons is mani- 
festly the same, and thus for each descent of either piston a 
volume of air equal to that of either barrel is expelled from the 
-machine. 

Let V be the volume of the receiver pipes and one barrel, V, 
the capacity of either barrel, and p„ the density of the air in the 
machine after the Tith stroke. Now the quantity of air in the 
machine after the nth stroke is p„V, and by the (n-\-l)th stroke 
the volume V,, or quantity p n V 1? is expelled. There will re- 
main then in the machine after the (n+l)th stroke the quan- 
tity p„V— p n V l9 and this being diffused through the space V, 
we have, by (155) and (156), 

P^V^nV-p^^V-V,), 

V \ 
or P»+x=P»(l— y)' 

If p be the initial density of the air, by making n equal to 
0, 1, 2, &c, successively, we have 

U 



306 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 

V \ 

v\ V V 

P 2 =PAl—y-)=Po{l—f) 

v y 

It is obvious that p„ can never become zero as long as n is 
finite, and therefore, even theoretically, perfect exhaustion is 
impossible. 

CONDENSER. 

443. If in the preceding figure the valves V, V, v, v', were 
made to open downward, it would represent a condenser. By 
each descent of the piston a volume of air equal to that of the 
barrel will be forced into R, and will by its reaction close the 
valves v and v' 9 and be retained there. 

To find the density of the air in the receiver after n strokes 
of the piston, let p n and p„ +1 be the densities of the air in R after 
the nth. and (n+l)th strokes, V the volume of the receiver R 
and the pipe ac, V , that of either barrel, and p the density of 
the external air. Then 

p n+1 Y= Pn Y+ Po V 1 . 
V, 

•*. P„+l = P« + P -y • 

If n be made successively equal to 0, 1, 2, 3, &c, we have 

V 2V \ 

p2=Pi+Po-Y=Po( 1 +-y l ) 
ftVA 



CLEPSYDRA. 307 

CLEPSYDRA. 

444. The clepsydra, or water clock, is a contrivance for 
measuring time by the descent of the surface of a fluid which 
flows through a small aperture in the base of the vessel con- 
taining it. 

Suppose the vessel a prism. It is required to determine 
what scale must be marked on its side, that the coincidence of 
the descending surface with the successive lines of division 
may mark equal successive intervals of time. Let a be the 
altitude of the prism, x the distance of the surface from the 
base of the vessel at the end of the time t, from the beginning 
of the motion. 

Equation (172) gives, using x instead of h— x, 

x-a g-l+^gif. 

Let &x and At be corresponding increments of x and t; then 

x+Ax=a-^^(t+At)+^(t+M)\ 
Subtracting the preceding equation from this, and we have 

The time t is in seconds, and, to determine the divisions cor- 
responding to successive minutes of time, put At=6Q", and give 
to t the successive values 0, 60", 120", &c. 

In a vessel of the form supposed in Ex. 4, Art. 433, the vert- 
ical distance of the successive divisions will all be equal, or 
the surface will descend equal distances in equal times. 



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